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Show that $A_4$ (which has order $12$) has exactly three elements of order $2$.

Additional information: $A_4$ denotes the set of even permutations in $S_4$. $S_4$ is defined as all of the permutations of size 4. $|S_n| = n!$ and an earlier exercise in the book I'm using (Hungerford, Abstract Algebra) shows that $|A_n| = n!/2$. There's no hint in the book and the exercise comes from Chapter 8.1 Exercise 44.

Edit: I've recently begun to learn about finding isomorphisms to a certain group using multiplication tables along with the knowledge of cosets (right-cosets specifically) and Lagrange's Theorem.

I think that the way to solve this problem might be to consider the elements of $A_4$ with orders 2,3 or 6. In general I'm feeling a bit lost in this field so I would really appreciate if someone had a good explanation.

Ludwwwig
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  • What definition of $A_4$ are you working with? Do you have a set of generators and relations? – Sammy Black Mar 03 '14 at 22:56
  • To get worthwhile help, you should say what you already know, and what you’ve tried. Many of us can hand you the answer on a plate, but that will not be any help to you. – Lubin Mar 03 '14 at 23:00
  • Here's a link so that you can see the Cayley table: it shows you flat out which have order 2: http://acunix.wheatonma.edu/jsklensk/Abstract_Fall08/classwork/november/nov3-inclass.pdf – Eleven-Eleven Mar 03 '14 at 23:00

2 Answers2

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clearly the $2-$cycle doesn't lie in $A_4$ (for some basic reference see the Wikipedia Article about this fact). So what are the elements of order 2 in here?

A few facts: --try to prove or recall--

1)the order of the product of two disjoint cycles is the lcm of their orders.

2) two - cycles have order 2, but they don't lie in A_4. But you have 4 different element to permute, so using the rule of signs for permutations you can conclude

3) there are nice books which treat these properties and basics fact very well (and it is important to understand them very well), for an introductory level you can start reading "Humpreys - A course in Group Theory", and when you feel more confident you can start trying to solve the exercise in "Rotman - An Introduction to the Theory of Groups". Maybe you can find an answer to my old question $S_n$ cannot be imbedded in $A_{n+1}$ for every $n$, answer I'm very interested to see :)

Community
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Riccardo
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  • Thank you for your answer. I've realized, just as you point out, that the 2-cycles are their own permutations but since they are odd they do not belong in A_4. Do you have a good method on listing the elements of A_4? If I understand your answer correctly, you're suggesting that I should find 4 elements and use the rule of signs. How should I find these elements and what's the rule of signs? – Ludwwwig Mar 04 '14 at 07:19
  • First of all, in this case I spoken about elements referring to the elements of the set,(an element of $A_4$ is a permutation over such set). It is irrelevant what this set is, it has four element. the rule of signs is basically the fact you have a group morphism "sign" from $S_n$ to $C_2$ (the cyclic group with two elements), so $sign(ab)=sign(a)sign(b)$. $A_n$ is the kernel of such morphism (and so it's a normal subgroup of $S_n$) – Riccardo Mar 04 '14 at 07:42
  • (Calling the element of $C_2$ $1,a$ with $a^2=1$ -obviously it is the only way to make a set with two element a group) So if you have two odd permutations $\varphi$ and $\pi$, then $sign(\varphi)=a$ , $sign(\pi)=a$ and by the fact that $sign$ is a morphism $sign(\varphi \pi) =sign(\varphi)sign(\pi)=a^2=1$ so the product is a element of the kernel of $sign$ and so it belongs to $A_n$. – Riccardo Mar 04 '14 at 07:51
  • @Ludwwwig and lastly, you need an even permutation of order two. You have the two-cycles whose order is $2$, but if you could multiplying them by another odd disjoint permutation of order two, you would have an even permutation of order the lcm of $2$ and $2$, so $2$. Here you need the fact that your base set has at least $4$ different elements. – Riccardo Mar 04 '14 at 07:54
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we should create elements in the form of $(ab)(cd) \implies {4 \choose 2} \cdot \frac{1}{2}=3 $

Why did we divided by $2$? Since $(ab)(cd)=(cd)(ab)$

Riccardo
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mesel
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