How to find the order of a symmetric group S4?How many transvections are in a maximal unipotent subgroup of a...
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How to find the order of a symmetric group S4?
How many transvections are in a maximal unipotent subgroup of a general linear group?Why is this set of $2 times 2$ matrices of real numbers a nonabelian group under multiplication?Cyclic subgroups and generatorsIs it a field or not?Group theory with matricesElements of order three in $GL_3(2)$The identity element in a sub groupCalculate the order of the following group elements.Why is this representation completely reducible?Classifying elements of finite and infinite order in $GL_{2}(mathbb{R})$
$begingroup$
Let $omega = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 3 & 2 & 4 & 1 end{smallmatrix}bigr)$
and
$tau = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 2 & 3 & 4 & 1 end{smallmatrix}bigr)$, both elements of $S_4$.
How exactly do I compute the orders of $omega$ and $tau$ ?
My professor said $ |omega|= 3$ and$ |tau|= 4$.
But he did not explain how to get these answers, also why are they both not just 4? I thought order was the number of elements in G, in this case they both have 4.
Does $ |omega|= 3$ because $omega(2)=2$, so it sends itself to itself?
Sorry, if this question seems dumb I just don't fully understand where the answer is coming from.
abstract-algebra group-theory permutations
$endgroup$
add a comment |
$begingroup$
Let $omega = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 3 & 2 & 4 & 1 end{smallmatrix}bigr)$
and
$tau = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 2 & 3 & 4 & 1 end{smallmatrix}bigr)$, both elements of $S_4$.
How exactly do I compute the orders of $omega$ and $tau$ ?
My professor said $ |omega|= 3$ and$ |tau|= 4$.
But he did not explain how to get these answers, also why are they both not just 4? I thought order was the number of elements in G, in this case they both have 4.
Does $ |omega|= 3$ because $omega(2)=2$, so it sends itself to itself?
Sorry, if this question seems dumb I just don't fully understand where the answer is coming from.
abstract-algebra group-theory permutations
$endgroup$
1
$begingroup$
Try compute $omega^3$, $tau^2$, $tau^4$ to get the order of these two elements.
$endgroup$
– Kemono Chen
4 hours ago
$begingroup$
It's not exactly about the number of fixed points. For example, $sigma= bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 2 &1 & 4 & 3 end{smallmatrix}bigr)$ has order 2. Can you see why, now that the correct definition of "order" in this context has been explained?
$endgroup$
– Robert Shore
4 hours ago
add a comment |
$begingroup$
Let $omega = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 3 & 2 & 4 & 1 end{smallmatrix}bigr)$
and
$tau = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 2 & 3 & 4 & 1 end{smallmatrix}bigr)$, both elements of $S_4$.
How exactly do I compute the orders of $omega$ and $tau$ ?
My professor said $ |omega|= 3$ and$ |tau|= 4$.
But he did not explain how to get these answers, also why are they both not just 4? I thought order was the number of elements in G, in this case they both have 4.
Does $ |omega|= 3$ because $omega(2)=2$, so it sends itself to itself?
Sorry, if this question seems dumb I just don't fully understand where the answer is coming from.
abstract-algebra group-theory permutations
$endgroup$
Let $omega = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 3 & 2 & 4 & 1 end{smallmatrix}bigr)$
and
$tau = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 2 & 3 & 4 & 1 end{smallmatrix}bigr)$, both elements of $S_4$.
How exactly do I compute the orders of $omega$ and $tau$ ?
My professor said $ |omega|= 3$ and$ |tau|= 4$.
But he did not explain how to get these answers, also why are they both not just 4? I thought order was the number of elements in G, in this case they both have 4.
Does $ |omega|= 3$ because $omega(2)=2$, so it sends itself to itself?
Sorry, if this question seems dumb I just don't fully understand where the answer is coming from.
abstract-algebra group-theory permutations
abstract-algebra group-theory permutations
edited 4 hours ago
Kemono Chen
3,1641844
3,1641844
asked 5 hours ago
User100290392039User100290392039
435
435
1
$begingroup$
Try compute $omega^3$, $tau^2$, $tau^4$ to get the order of these two elements.
$endgroup$
– Kemono Chen
4 hours ago
$begingroup$
It's not exactly about the number of fixed points. For example, $sigma= bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 2 &1 & 4 & 3 end{smallmatrix}bigr)$ has order 2. Can you see why, now that the correct definition of "order" in this context has been explained?
$endgroup$
– Robert Shore
4 hours ago
add a comment |
1
$begingroup$
Try compute $omega^3$, $tau^2$, $tau^4$ to get the order of these two elements.
$endgroup$
– Kemono Chen
4 hours ago
$begingroup$
It's not exactly about the number of fixed points. For example, $sigma= bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 2 &1 & 4 & 3 end{smallmatrix}bigr)$ has order 2. Can you see why, now that the correct definition of "order" in this context has been explained?
$endgroup$
– Robert Shore
4 hours ago
1
1
$begingroup$
Try compute $omega^3$, $tau^2$, $tau^4$ to get the order of these two elements.
$endgroup$
– Kemono Chen
4 hours ago
$begingroup$
Try compute $omega^3$, $tau^2$, $tau^4$ to get the order of these two elements.
$endgroup$
– Kemono Chen
4 hours ago
$begingroup$
It's not exactly about the number of fixed points. For example, $sigma= bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 2 &1 & 4 & 3 end{smallmatrix}bigr)$ has order 2. Can you see why, now that the correct definition of "order" in this context has been explained?
$endgroup$
– Robert Shore
4 hours ago
$begingroup$
It's not exactly about the number of fixed points. For example, $sigma= bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 2 &1 & 4 & 3 end{smallmatrix}bigr)$ has order 2. Can you see why, now that the correct definition of "order" in this context has been explained?
$endgroup$
– Robert Shore
4 hours ago
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
The order of a group is not the same thing as the order of an element. There is a connection, but don't worry about that yet. The order of a group, written $|G|$, is the number of elements it has. The order of an element in a group - say $g in G$, is the smallest $n$ such $g^n = e$ where $e$ is the identity element of the group $G$. In this case we also write $|g| = n$.
You are not being asked to find the order of a group. $tau$ and $omega$ are elements of a group and you are being asked to find the order of $tau$ and $omega$ as elements of $S_4$. Elements of $S_4$, such as $tau$ and $omega$, are bijections (functions) from the set ${1,2,3,4}$ to the set ${1,2,3,4}$. To find their order you need to investigate how many times you need to repeatedly apply the function in order to get the identity function. It doesn't really make sense to take about the order of $tau$ and $omega$ based on how many elements they have as you did in your original post, however we do have this.
It does turn out that if the order of an element is $n$, say $|g| = n$, then the cyclic subgroup generated by $g$ will have $n$ elements. So I guess if you wanted to find the orders of $tau$ and $omega$ you could go count the elements of their cyclic subgroups but this seems superfluous.
$endgroup$
$begingroup$
If we want to find the order of $S_4$, it should be $4!$, right?
$endgroup$
– manooooh
4 hours ago
2
$begingroup$
Yes, but if you read the question you will realize that the title of the question is not really what the OP is asking.
$endgroup$
– Prince M
4 hours ago
add a comment |
$begingroup$
We first need to know that when talking about $S_4$, we are talking about the set of all bijections from the set ${1,2,3,4}$ to itself.
So, the elements of the set $S_4$ are functions. An example of a function in $S_4$ is the function $f: {1,2,3,4} to {1,2,3,4}$ given by:
$$1 mapsto 2, 2mapsto 3, 3mapsto 4, 4 mapsto 1 $$
However, this is incredibly tedious to write out. Instead, we will use matrices to represent functions in $S_4$. In other words, we will let:
$$f := bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 2 & 3 & 4 & 1 end{smallmatrix}bigr)$$
where an element in row 1 of the matrix is mapped to the element below it in row 2 of the matrix by $f$.
Now, when we are talking about order, we can talk about 2 things: the order of a group, and the order of an element of the group.
The order of a group is the number of elements in that set, but we can also pick out an element from that group (in this case the group is $S_4$ and we're picking out an element from $S_4$) and we can ask what is the order of the element. The order of an element is the smallest positive number $n$ such that $x^n = 1$ where $x$ is the element of the group we are considering, and $1$ is the identity element in the group.
In your case, when your professor says that $|omega| = 3$, what he means is that if you compose the function $omega in S_4$ with itself 3 times, you will get back the identity element of $S_4$. The identity element of $S_4$ is the identity function on ${1,2,3,4}$, i.e. the function:
$$i = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 1 & 2 & 3 & 4 end{smallmatrix}bigr)$$
For instance, although $omega = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 3 & 2 & 4 & 1 end{smallmatrix}bigr)$, we see that $omega^2 = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 4 & 2 & 1 & 3 end{smallmatrix}bigr)$, and $omega^3 = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 1 & 2 & 3 & 4 end{smallmatrix}bigr) = i$
$endgroup$
add a comment |
$begingroup$
In "cycle notation ", $omega=(134)$ is a $3$-cycle, and $tau=(1234)$ is a $4$-cycle.
It happens that $n$-cycles always have order $n$.
$endgroup$
add a comment |
$begingroup$
We may always write a permutation as 1 or more cyclic permutations:
For example, for $omega = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 3 & 2 & 4 & 1 end{smallmatrix}bigr)$ you may see, that
$$1 mapsto 3 mapsto 4 mapsto 1$$
and
$$2 mapsto 2$$
so we obtained cyclic permutations of 3 elements and 1 element, respective.
Using the notation for cyclic permutations, the first one is
$$(1 3 4)$$
and the second one is simply $$(2)$$
As the result, we may write
$$omega = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 3 & 2 & 4 & 1 end{smallmatrix}bigr)= (1 3 4) (2)$$
Now, the first, 3-element cyclic permutation has the order 3 (its "cycle" must turn 3-times $120^o$, to return to its initial position), the second 1 (order of identity). Their least common multiple is 3, so
$$ord(omega) = 3$$
For $tau = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 2 & 3 & 4 & 1 end{smallmatrix}bigr)$ is the situation very simple, because it is a cyclic one :
$$1 mapsto 2 mapsto 3 mapsto 4 mapsto 1$$
so we may write
$$tau = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 2 & 3 & 4 & 1 end{smallmatrix}bigr) = (1 2 3 4)$$
and
$$ord(tau) = 4$$
$endgroup$
add a comment |
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4 Answers
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oldest
votes
4 Answers
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$begingroup$
The order of a group is not the same thing as the order of an element. There is a connection, but don't worry about that yet. The order of a group, written $|G|$, is the number of elements it has. The order of an element in a group - say $g in G$, is the smallest $n$ such $g^n = e$ where $e$ is the identity element of the group $G$. In this case we also write $|g| = n$.
You are not being asked to find the order of a group. $tau$ and $omega$ are elements of a group and you are being asked to find the order of $tau$ and $omega$ as elements of $S_4$. Elements of $S_4$, such as $tau$ and $omega$, are bijections (functions) from the set ${1,2,3,4}$ to the set ${1,2,3,4}$. To find their order you need to investigate how many times you need to repeatedly apply the function in order to get the identity function. It doesn't really make sense to take about the order of $tau$ and $omega$ based on how many elements they have as you did in your original post, however we do have this.
It does turn out that if the order of an element is $n$, say $|g| = n$, then the cyclic subgroup generated by $g$ will have $n$ elements. So I guess if you wanted to find the orders of $tau$ and $omega$ you could go count the elements of their cyclic subgroups but this seems superfluous.
$endgroup$
$begingroup$
If we want to find the order of $S_4$, it should be $4!$, right?
$endgroup$
– manooooh
4 hours ago
2
$begingroup$
Yes, but if you read the question you will realize that the title of the question is not really what the OP is asking.
$endgroup$
– Prince M
4 hours ago
add a comment |
$begingroup$
The order of a group is not the same thing as the order of an element. There is a connection, but don't worry about that yet. The order of a group, written $|G|$, is the number of elements it has. The order of an element in a group - say $g in G$, is the smallest $n$ such $g^n = e$ where $e$ is the identity element of the group $G$. In this case we also write $|g| = n$.
You are not being asked to find the order of a group. $tau$ and $omega$ are elements of a group and you are being asked to find the order of $tau$ and $omega$ as elements of $S_4$. Elements of $S_4$, such as $tau$ and $omega$, are bijections (functions) from the set ${1,2,3,4}$ to the set ${1,2,3,4}$. To find their order you need to investigate how many times you need to repeatedly apply the function in order to get the identity function. It doesn't really make sense to take about the order of $tau$ and $omega$ based on how many elements they have as you did in your original post, however we do have this.
It does turn out that if the order of an element is $n$, say $|g| = n$, then the cyclic subgroup generated by $g$ will have $n$ elements. So I guess if you wanted to find the orders of $tau$ and $omega$ you could go count the elements of their cyclic subgroups but this seems superfluous.
$endgroup$
$begingroup$
If we want to find the order of $S_4$, it should be $4!$, right?
$endgroup$
– manooooh
4 hours ago
2
$begingroup$
Yes, but if you read the question you will realize that the title of the question is not really what the OP is asking.
$endgroup$
– Prince M
4 hours ago
add a comment |
$begingroup$
The order of a group is not the same thing as the order of an element. There is a connection, but don't worry about that yet. The order of a group, written $|G|$, is the number of elements it has. The order of an element in a group - say $g in G$, is the smallest $n$ such $g^n = e$ where $e$ is the identity element of the group $G$. In this case we also write $|g| = n$.
You are not being asked to find the order of a group. $tau$ and $omega$ are elements of a group and you are being asked to find the order of $tau$ and $omega$ as elements of $S_4$. Elements of $S_4$, such as $tau$ and $omega$, are bijections (functions) from the set ${1,2,3,4}$ to the set ${1,2,3,4}$. To find their order you need to investigate how many times you need to repeatedly apply the function in order to get the identity function. It doesn't really make sense to take about the order of $tau$ and $omega$ based on how many elements they have as you did in your original post, however we do have this.
It does turn out that if the order of an element is $n$, say $|g| = n$, then the cyclic subgroup generated by $g$ will have $n$ elements. So I guess if you wanted to find the orders of $tau$ and $omega$ you could go count the elements of their cyclic subgroups but this seems superfluous.
$endgroup$
The order of a group is not the same thing as the order of an element. There is a connection, but don't worry about that yet. The order of a group, written $|G|$, is the number of elements it has. The order of an element in a group - say $g in G$, is the smallest $n$ such $g^n = e$ where $e$ is the identity element of the group $G$. In this case we also write $|g| = n$.
You are not being asked to find the order of a group. $tau$ and $omega$ are elements of a group and you are being asked to find the order of $tau$ and $omega$ as elements of $S_4$. Elements of $S_4$, such as $tau$ and $omega$, are bijections (functions) from the set ${1,2,3,4}$ to the set ${1,2,3,4}$. To find their order you need to investigate how many times you need to repeatedly apply the function in order to get the identity function. It doesn't really make sense to take about the order of $tau$ and $omega$ based on how many elements they have as you did in your original post, however we do have this.
It does turn out that if the order of an element is $n$, say $|g| = n$, then the cyclic subgroup generated by $g$ will have $n$ elements. So I guess if you wanted to find the orders of $tau$ and $omega$ you could go count the elements of their cyclic subgroups but this seems superfluous.
edited 4 hours ago
answered 4 hours ago
Prince MPrince M
2,0531521
2,0531521
$begingroup$
If we want to find the order of $S_4$, it should be $4!$, right?
$endgroup$
– manooooh
4 hours ago
2
$begingroup$
Yes, but if you read the question you will realize that the title of the question is not really what the OP is asking.
$endgroup$
– Prince M
4 hours ago
add a comment |
$begingroup$
If we want to find the order of $S_4$, it should be $4!$, right?
$endgroup$
– manooooh
4 hours ago
2
$begingroup$
Yes, but if you read the question you will realize that the title of the question is not really what the OP is asking.
$endgroup$
– Prince M
4 hours ago
$begingroup$
If we want to find the order of $S_4$, it should be $4!$, right?
$endgroup$
– manooooh
4 hours ago
$begingroup$
If we want to find the order of $S_4$, it should be $4!$, right?
$endgroup$
– manooooh
4 hours ago
2
2
$begingroup$
Yes, but if you read the question you will realize that the title of the question is not really what the OP is asking.
$endgroup$
– Prince M
4 hours ago
$begingroup$
Yes, but if you read the question you will realize that the title of the question is not really what the OP is asking.
$endgroup$
– Prince M
4 hours ago
add a comment |
$begingroup$
We first need to know that when talking about $S_4$, we are talking about the set of all bijections from the set ${1,2,3,4}$ to itself.
So, the elements of the set $S_4$ are functions. An example of a function in $S_4$ is the function $f: {1,2,3,4} to {1,2,3,4}$ given by:
$$1 mapsto 2, 2mapsto 3, 3mapsto 4, 4 mapsto 1 $$
However, this is incredibly tedious to write out. Instead, we will use matrices to represent functions in $S_4$. In other words, we will let:
$$f := bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 2 & 3 & 4 & 1 end{smallmatrix}bigr)$$
where an element in row 1 of the matrix is mapped to the element below it in row 2 of the matrix by $f$.
Now, when we are talking about order, we can talk about 2 things: the order of a group, and the order of an element of the group.
The order of a group is the number of elements in that set, but we can also pick out an element from that group (in this case the group is $S_4$ and we're picking out an element from $S_4$) and we can ask what is the order of the element. The order of an element is the smallest positive number $n$ such that $x^n = 1$ where $x$ is the element of the group we are considering, and $1$ is the identity element in the group.
In your case, when your professor says that $|omega| = 3$, what he means is that if you compose the function $omega in S_4$ with itself 3 times, you will get back the identity element of $S_4$. The identity element of $S_4$ is the identity function on ${1,2,3,4}$, i.e. the function:
$$i = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 1 & 2 & 3 & 4 end{smallmatrix}bigr)$$
For instance, although $omega = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 3 & 2 & 4 & 1 end{smallmatrix}bigr)$, we see that $omega^2 = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 4 & 2 & 1 & 3 end{smallmatrix}bigr)$, and $omega^3 = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 1 & 2 & 3 & 4 end{smallmatrix}bigr) = i$
$endgroup$
add a comment |
$begingroup$
We first need to know that when talking about $S_4$, we are talking about the set of all bijections from the set ${1,2,3,4}$ to itself.
So, the elements of the set $S_4$ are functions. An example of a function in $S_4$ is the function $f: {1,2,3,4} to {1,2,3,4}$ given by:
$$1 mapsto 2, 2mapsto 3, 3mapsto 4, 4 mapsto 1 $$
However, this is incredibly tedious to write out. Instead, we will use matrices to represent functions in $S_4$. In other words, we will let:
$$f := bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 2 & 3 & 4 & 1 end{smallmatrix}bigr)$$
where an element in row 1 of the matrix is mapped to the element below it in row 2 of the matrix by $f$.
Now, when we are talking about order, we can talk about 2 things: the order of a group, and the order of an element of the group.
The order of a group is the number of elements in that set, but we can also pick out an element from that group (in this case the group is $S_4$ and we're picking out an element from $S_4$) and we can ask what is the order of the element. The order of an element is the smallest positive number $n$ such that $x^n = 1$ where $x$ is the element of the group we are considering, and $1$ is the identity element in the group.
In your case, when your professor says that $|omega| = 3$, what he means is that if you compose the function $omega in S_4$ with itself 3 times, you will get back the identity element of $S_4$. The identity element of $S_4$ is the identity function on ${1,2,3,4}$, i.e. the function:
$$i = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 1 & 2 & 3 & 4 end{smallmatrix}bigr)$$
For instance, although $omega = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 3 & 2 & 4 & 1 end{smallmatrix}bigr)$, we see that $omega^2 = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 4 & 2 & 1 & 3 end{smallmatrix}bigr)$, and $omega^3 = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 1 & 2 & 3 & 4 end{smallmatrix}bigr) = i$
$endgroup$
add a comment |
$begingroup$
We first need to know that when talking about $S_4$, we are talking about the set of all bijections from the set ${1,2,3,4}$ to itself.
So, the elements of the set $S_4$ are functions. An example of a function in $S_4$ is the function $f: {1,2,3,4} to {1,2,3,4}$ given by:
$$1 mapsto 2, 2mapsto 3, 3mapsto 4, 4 mapsto 1 $$
However, this is incredibly tedious to write out. Instead, we will use matrices to represent functions in $S_4$. In other words, we will let:
$$f := bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 2 & 3 & 4 & 1 end{smallmatrix}bigr)$$
where an element in row 1 of the matrix is mapped to the element below it in row 2 of the matrix by $f$.
Now, when we are talking about order, we can talk about 2 things: the order of a group, and the order of an element of the group.
The order of a group is the number of elements in that set, but we can also pick out an element from that group (in this case the group is $S_4$ and we're picking out an element from $S_4$) and we can ask what is the order of the element. The order of an element is the smallest positive number $n$ such that $x^n = 1$ where $x$ is the element of the group we are considering, and $1$ is the identity element in the group.
In your case, when your professor says that $|omega| = 3$, what he means is that if you compose the function $omega in S_4$ with itself 3 times, you will get back the identity element of $S_4$. The identity element of $S_4$ is the identity function on ${1,2,3,4}$, i.e. the function:
$$i = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 1 & 2 & 3 & 4 end{smallmatrix}bigr)$$
For instance, although $omega = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 3 & 2 & 4 & 1 end{smallmatrix}bigr)$, we see that $omega^2 = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 4 & 2 & 1 & 3 end{smallmatrix}bigr)$, and $omega^3 = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 1 & 2 & 3 & 4 end{smallmatrix}bigr) = i$
$endgroup$
We first need to know that when talking about $S_4$, we are talking about the set of all bijections from the set ${1,2,3,4}$ to itself.
So, the elements of the set $S_4$ are functions. An example of a function in $S_4$ is the function $f: {1,2,3,4} to {1,2,3,4}$ given by:
$$1 mapsto 2, 2mapsto 3, 3mapsto 4, 4 mapsto 1 $$
However, this is incredibly tedious to write out. Instead, we will use matrices to represent functions in $S_4$. In other words, we will let:
$$f := bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 2 & 3 & 4 & 1 end{smallmatrix}bigr)$$
where an element in row 1 of the matrix is mapped to the element below it in row 2 of the matrix by $f$.
Now, when we are talking about order, we can talk about 2 things: the order of a group, and the order of an element of the group.
The order of a group is the number of elements in that set, but we can also pick out an element from that group (in this case the group is $S_4$ and we're picking out an element from $S_4$) and we can ask what is the order of the element. The order of an element is the smallest positive number $n$ such that $x^n = 1$ where $x$ is the element of the group we are considering, and $1$ is the identity element in the group.
In your case, when your professor says that $|omega| = 3$, what he means is that if you compose the function $omega in S_4$ with itself 3 times, you will get back the identity element of $S_4$. The identity element of $S_4$ is the identity function on ${1,2,3,4}$, i.e. the function:
$$i = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 1 & 2 & 3 & 4 end{smallmatrix}bigr)$$
For instance, although $omega = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 3 & 2 & 4 & 1 end{smallmatrix}bigr)$, we see that $omega^2 = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 4 & 2 & 1 & 3 end{smallmatrix}bigr)$, and $omega^3 = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 1 & 2 & 3 & 4 end{smallmatrix}bigr) = i$
edited 4 hours ago
answered 4 hours ago
user516079user516079
363210
363210
add a comment |
add a comment |
$begingroup$
In "cycle notation ", $omega=(134)$ is a $3$-cycle, and $tau=(1234)$ is a $4$-cycle.
It happens that $n$-cycles always have order $n$.
$endgroup$
add a comment |
$begingroup$
In "cycle notation ", $omega=(134)$ is a $3$-cycle, and $tau=(1234)$ is a $4$-cycle.
It happens that $n$-cycles always have order $n$.
$endgroup$
add a comment |
$begingroup$
In "cycle notation ", $omega=(134)$ is a $3$-cycle, and $tau=(1234)$ is a $4$-cycle.
It happens that $n$-cycles always have order $n$.
$endgroup$
In "cycle notation ", $omega=(134)$ is a $3$-cycle, and $tau=(1234)$ is a $4$-cycle.
It happens that $n$-cycles always have order $n$.
answered 3 hours ago
Chris CusterChris Custer
13.9k3827
13.9k3827
add a comment |
add a comment |
$begingroup$
We may always write a permutation as 1 or more cyclic permutations:
For example, for $omega = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 3 & 2 & 4 & 1 end{smallmatrix}bigr)$ you may see, that
$$1 mapsto 3 mapsto 4 mapsto 1$$
and
$$2 mapsto 2$$
so we obtained cyclic permutations of 3 elements and 1 element, respective.
Using the notation for cyclic permutations, the first one is
$$(1 3 4)$$
and the second one is simply $$(2)$$
As the result, we may write
$$omega = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 3 & 2 & 4 & 1 end{smallmatrix}bigr)= (1 3 4) (2)$$
Now, the first, 3-element cyclic permutation has the order 3 (its "cycle" must turn 3-times $120^o$, to return to its initial position), the second 1 (order of identity). Their least common multiple is 3, so
$$ord(omega) = 3$$
For $tau = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 2 & 3 & 4 & 1 end{smallmatrix}bigr)$ is the situation very simple, because it is a cyclic one :
$$1 mapsto 2 mapsto 3 mapsto 4 mapsto 1$$
so we may write
$$tau = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 2 & 3 & 4 & 1 end{smallmatrix}bigr) = (1 2 3 4)$$
and
$$ord(tau) = 4$$
$endgroup$
add a comment |
$begingroup$
We may always write a permutation as 1 or more cyclic permutations:
For example, for $omega = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 3 & 2 & 4 & 1 end{smallmatrix}bigr)$ you may see, that
$$1 mapsto 3 mapsto 4 mapsto 1$$
and
$$2 mapsto 2$$
so we obtained cyclic permutations of 3 elements and 1 element, respective.
Using the notation for cyclic permutations, the first one is
$$(1 3 4)$$
and the second one is simply $$(2)$$
As the result, we may write
$$omega = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 3 & 2 & 4 & 1 end{smallmatrix}bigr)= (1 3 4) (2)$$
Now, the first, 3-element cyclic permutation has the order 3 (its "cycle" must turn 3-times $120^o$, to return to its initial position), the second 1 (order of identity). Their least common multiple is 3, so
$$ord(omega) = 3$$
For $tau = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 2 & 3 & 4 & 1 end{smallmatrix}bigr)$ is the situation very simple, because it is a cyclic one :
$$1 mapsto 2 mapsto 3 mapsto 4 mapsto 1$$
so we may write
$$tau = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 2 & 3 & 4 & 1 end{smallmatrix}bigr) = (1 2 3 4)$$
and
$$ord(tau) = 4$$
$endgroup$
add a comment |
$begingroup$
We may always write a permutation as 1 or more cyclic permutations:
For example, for $omega = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 3 & 2 & 4 & 1 end{smallmatrix}bigr)$ you may see, that
$$1 mapsto 3 mapsto 4 mapsto 1$$
and
$$2 mapsto 2$$
so we obtained cyclic permutations of 3 elements and 1 element, respective.
Using the notation for cyclic permutations, the first one is
$$(1 3 4)$$
and the second one is simply $$(2)$$
As the result, we may write
$$omega = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 3 & 2 & 4 & 1 end{smallmatrix}bigr)= (1 3 4) (2)$$
Now, the first, 3-element cyclic permutation has the order 3 (its "cycle" must turn 3-times $120^o$, to return to its initial position), the second 1 (order of identity). Their least common multiple is 3, so
$$ord(omega) = 3$$
For $tau = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 2 & 3 & 4 & 1 end{smallmatrix}bigr)$ is the situation very simple, because it is a cyclic one :
$$1 mapsto 2 mapsto 3 mapsto 4 mapsto 1$$
so we may write
$$tau = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 2 & 3 & 4 & 1 end{smallmatrix}bigr) = (1 2 3 4)$$
and
$$ord(tau) = 4$$
$endgroup$
We may always write a permutation as 1 or more cyclic permutations:
For example, for $omega = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 3 & 2 & 4 & 1 end{smallmatrix}bigr)$ you may see, that
$$1 mapsto 3 mapsto 4 mapsto 1$$
and
$$2 mapsto 2$$
so we obtained cyclic permutations of 3 elements and 1 element, respective.
Using the notation for cyclic permutations, the first one is
$$(1 3 4)$$
and the second one is simply $$(2)$$
As the result, we may write
$$omega = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 3 & 2 & 4 & 1 end{smallmatrix}bigr)= (1 3 4) (2)$$
Now, the first, 3-element cyclic permutation has the order 3 (its "cycle" must turn 3-times $120^o$, to return to its initial position), the second 1 (order of identity). Their least common multiple is 3, so
$$ord(omega) = 3$$
For $tau = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 2 & 3 & 4 & 1 end{smallmatrix}bigr)$ is the situation very simple, because it is a cyclic one :
$$1 mapsto 2 mapsto 3 mapsto 4 mapsto 1$$
so we may write
$$tau = bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 2 & 3 & 4 & 1 end{smallmatrix}bigr) = (1 2 3 4)$$
and
$$ord(tau) = 4$$
edited 1 hour ago
answered 1 hour ago
MarianDMarianD
5381511
5381511
add a comment |
add a comment |
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$begingroup$
Try compute $omega^3$, $tau^2$, $tau^4$ to get the order of these two elements.
$endgroup$
– Kemono Chen
4 hours ago
$begingroup$
It's not exactly about the number of fixed points. For example, $sigma= bigl(begin{smallmatrix}1 & 2 & 3 & 4 \ 2 &1 & 4 & 3 end{smallmatrix}bigr)$ has order 2. Can you see why, now that the correct definition of "order" in this context has been explained?
$endgroup$
– Robert Shore
4 hours ago