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Thread: Isomorphism problem

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    Isomorphism problem

    I'm doing a problem in my Algebra text. We are given 2 3-cycles in $\displaystyle S_n$ , x and y, and told that <x, y> is isomorphic to one of the following: $\displaystyle \mathbb{Z}_3 \times \mathbb{Z}_3 $, $\displaystyle A_4$, $\displaystyle A_5$ and $\displaystyle \mathbb{Z}_3$.

    Since we can reorder the permutations at will I have chosen the example x = (23145) and y = (12453) to work out a more general method. (I am ignoring the elements that aren't affected by the permutation.)

    To start with we have the group elements for <x, y>: $\displaystyle 1, ~ x, ~ x^2, y, ~ y^2, ~ xy, ~ xy^2, ~ x^2y, ~ x^2y^2$

    Since the elements don't commute we have to include elements such as yx, etc. We will also have elements of the form $\displaystyle x \cdot yx$ which I can't find a way to simplify and $\displaystyle x \cdot xy = x^2y$ which is not a new element and thus can't be counted.

    I can certainly work out all the group elements and construct the whole thing, but that's just silly. My problem is this: I have constructed more than 10 elements in my list (actually a lot more) so <x, y> must be isomorphic to $\displaystyle A_5$ which has 60 elements, not to mention that it is the only group listed that can be non-Abelian. How do I count the number of elements in <x, y>?

    Thanks!

    -Dan
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    Re: Isomorphism problem

    Dan,
    I believe the result, but I don't understand your post. You are told that x and y are both 3 cycles and you have written x and y as 5 cycles?

    Here's some hints to get you started:

    As you point out, you can assume x=(123). Let y be a 3 cycle:

    1. Suppose y is disjoint from x. Then "easily" <x,y> is the direct product of 2 cyclic groups of order 3, viz. $Z_3\times Z_3$

    2. Suppose y has exactly 2 letters in common with x, say 1 and 2; you can assume the other letter of y is 4. Say y=(124). Then <x,y> is a subgroup H of $A_4$ such that 3 divides the order of H and the order of H is greater than 3. So the order of H is 6 or 12. Can order H be 6? No. Any group of order 6 has a normal subgroup of order 3 and so if $A_4$ contained a subgroup of order 6, it would contain a normal subgroup of order 3, contradiction. So the order of H is 12 and hence $H=A_4$.

    3. Suppose y has all letters in common with x. Easy. <x,y> is cyclic of order 3.

    4. "Hardest" case. Suppose y has exactly 1 letter in common with x, say 1. As in 2, you can assume y's other two letters are 4 and 5. Then <x,y> is $A_5$. I'll leave you here. If you need help, just say so.
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    Re: Isomorphism problem

    Quote Originally Posted by johng View Post
    Dan,
    I believe the result, but I don't understand your post. You are told that x and y are both 3 cycles and you have written x and y as 5 cycles?

    Here's some hints to get you started:

    As you point out, you can assume x=(123). Let y be a 3 cycle:

    1. Suppose y is disjoint from x. Then "easily" <x,y> is the direct product of 2 cyclic groups of order 3, viz. $Z_3\times Z_3$

    2. Suppose y has exactly 2 letters in common with x, say 1 and 2; you can assume the other letter of y is 4. Say y=(124). Then <x,y> is a subgroup H of $A_4$ such that 3 divides the order of H and the order of H is greater than 3. So the order of H is 6 or 12. Can order H be 6? No. Any group of order 6 has a normal subgroup of order 3 and so if $A_4$ contained a subgroup of order 6, it would contain a normal subgroup of order 3, contradiction. So the order of H is 12 and hence $H=A_4$.

    3. Suppose y has all letters in common with x. Easy. <x,y> is cyclic of order 3.

    4. "Hardest" case. Suppose y has exactly 1 letter in common with x, say 1. As in 2, you can assume y's other two letters are 4 and 5. Then <x,y> is $A_5$. I'll leave you here. If you need help, just say so.
    Sorry for any confusion. What I am trying to do is your "hardest case" with only one letter in common. I couldn't see how to do it without permutations on $\displaystyle S_5$ so I used letters up to 5. But x and y are each 3 cycle subgroups of what winds up being $\displaystyle A_5$.

    I got case 1 and case 3 and was thinking about how to do case 2. My method was to count the total number of elements in the group then compare elements...ie. if <x, y> contained a total of 12 elements then I could construct an isomorphism to $\displaystyle A_4$. Your way is much better. Obviously my programme doesn't work well with something as large as $\displaystyle A_5$.

    I'll give it a go and see what I can do from here.

    Thanks!

    -Dan
    Last edited by topsquark; Jun 2nd 2018 at 03:47 PM.
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    Re: Isomorphism problem

    Okay, I got 4). (Okay, I didn't actually prove that |<x, y>| couldn't be 45 but I got the rest.)

    I spent a fair amount of time on finishing out 2). I was trying to be as general as possible but I eventually had to do some constructions. How do we know that a group of order 6 contains a normal subgroup of order 3? I mean, it's kind of obvious when you think about it but I had to construct the group in order to prove it. Can it be proved or is it simply a fact that we can know by construction?

    -Dan
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    Re: Isomorphism problem

    Quote Originally Posted by topsquark View Post
    How do we know that a group of order 6 contains a normal subgroup of order 3?

    -Dan
    Theorem (Cauchy): If prime p divides the order of G then G contains an element (therefore a subgroup) of order p

    Fact: A subgroup of index 2 is normal

    We could also use the fact that there are, up to isomorphism, only two groups of order 6
    Last edited by Idea; Jun 12th 2018 at 12:22 AM.
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