# Thread: Isometric to symmetric group on...

1. ## Isometric to symmetric group on...

I have this problem:

Let Y = {a, b, c, d} and let X = Sym(Y)

(a) Find cyclic subgroups of X of order 4:

my answer: (a)(b)(c)(d), (a)(c)(cd), (a)(bc)(d), (a)(bcd). (Correct?)

This part I need a little help with:

(b) Show that X has subgroup L which is isomorphic to the symetric group on a set with 3 elements...

Not sure how to start.

2. I'm not entirely sure what you mean by (1)(2)(3)(4), is this just the identity permutation?

Notation aside, your answer is not correct. You need to take all permutations of order 4 and multiply them individually to find the subgroups generated by them. This will mean that every such group will contain an element of order 4. The subgroup you give contains the identity, two elements of order 2 and an element of order 3...You cannot have an element of order 3 as the order of an element must divide the order of a group. Here your subgroup (viewed as a group) has order 4, but 3 does not divide 4.

Now, as it happens all permutations of order 4 in $S_4$ have the same form (can you tell me what they look like?). So, take one of these elements and write down $\sigma$, $\sigma^2$, $\sigma^3$ (and you know that $\sigma^4=1$). This is one of your cyclic subgroups, the others look identical, just with a different permutation to start with.

For the second part, pick an element, say a'. What does the set of all permutation in $S_4$ which fix a' look like?

3. What I meant for (a) is have:

{a,b,c,d}, {a,b,d,c}, {a,c,b,d}, {a,c,d,b}, {a,d,c,b}, {a,d,b,c}

These are the 6 cyclic permutations of order 4 in X, correct?

I'm looking for the cyclic subgroups of X of order 4 though... Clearly I need to "bite the bullet" and go back to the start with this stuff... so little time.

Does $a=a, a^2=b, a^3=c$ .... argh, I feel like I'm inventing my own kind of math now.

So, take one of these elements and write down $\sigma, \sigma^2, \sigma^3$. This is one of your cyclic subgroups...
Could you actually write this out for one of the elements in terms of a, b, c, and d and I'll see if I can make sense of it?

P.S. Do you know where there are any good online lecture notes for group theory written in plane english as opposed the language used in many textbooks (my mathspeak is a little weak at the moment)?

4. Yeah, they're your 4 permutations. To work out your groups you just need to know how permutations are multiplied. Wikipedia seems to cover this well enough.

As for a book, Andrew Perry's book, Introduction to Abstract Algebra: Groups Rings and Fields' is a very good introduction to the subject of algebra on the whole, and for a website I either look up your favourite university and find out who is lecturing their introductory groups course. Then, go to their personal home page and they will probably have their notes there. If they don't, pick another university, and repeat...Also, I know MIT have a series of online lecture notes in pretty much all their subjects. You may want to check them out. However, I have never tried them so I can't really recommend them...

5. Originally Posted by Swlabr
Now, as it happens all permutations of order 4 in $S_4$ have the same form (can you tell me what they look like?). So, take one of these elements and write down $\sigma$, $\sigma^2$, $\sigma^3$ (and you know that $\sigma^4=1$). This is one of your cyclic subgroups, the others look identical, just with a different permutation to start with.
Does this make any sense? For example: { $a, a^2, a^3, | a^4=1$} only <a> generates the set.

So the cyclic subgroups of X of order 4 are <a>, <b>, <c>, and <d> ?

6. Originally Posted by MichaelMath
Does this make any sense? For example: { $a, a^2, a^3, | a^4=1$} only <a> generates the set.

So the cyclic subgroups of X of order 4 are <a>, <b>, <c>, and <d> ?
$a$ is not an element of your group. The elements of your group are the permutations, and the cyclic subgroups of order 4 are the subgroups generated by the permutations of order 4.

So, for example, take $\sigma = (1234)$. $\sigma$ has order 4, as you pointed out earlier. So one of your cyclic groups of order 4 will be,

$\{\sigma, \sigma^2, \sigma^3, id\}$

You not just need to calculate $\sigma^2$ and $\sigma^3$.

7. so,

$\sigma^2 = \sigma^2, (\sigma^2)^2 = \sigma^4=1$ ... doesn't generate the element.

$\sigma^3 = \sigma^3, (\sigma^3)^2 = \sigma^6= \sigma^2, (\sigma^3)^3=\sigma^9= \sigma$ ... generates the element.

So their are two cyclic subgroups for each element, the one you gave, and $\{\sigma^3, \sigma^2, \sigma, id\}$

Correct?

8. By the way, does sym(Y) consist of all 24 permutations of Y?

For the second part, pick an element, say a'. What does the set of all permutation in $S_4$ which fix `a' look like?
$L =\bigl\{ \{a,b,c,d\}, \{a,b,d,c\}, \{a,c,b,d\}, \{a,c,d,b\}, \{a,d,c,b\}, \{a,d,b,c\}\bigr\}$

Is this the subgroup I was looking for in part (b)?

9. Yes, Sym(Y), also denoted by $S_Y$, is the group of all permutations of the set $Y$ under the operation of composition of functions.

Now, you are asked to find the cyclic subgroups of $S_Y$ of order 4. To do this, you need to find a permutation of order 4. Can you tell me what these look like?

This element of order 4 will generate your cyclic subgroup.

I will demonstrate by finding a subgroup of order 3. So, note that $\sigma=(123)$ is a permutation of order 3 ( $\sigma^3=1$ but $\sigma^2 \neq 1$ and $\sigma \neq 1$).

So, our cyclic subgroup of order 3 will be $\{\sigma, \sigma^2, \sigma^3=1\}$. This is just the set $\{(123), (132), 1\}$ as $\sigma^2 = (123)(123) = (132)$.

You now just need to do the same, but with a permutation of order 4.