# About alternating group $A_4$

This is a simple exercise telling that $A_4$ cannot have a subgroup of order $6$. Here in my way:

Obviously, for any group $G$ and a subgroup $H$ of it with index $2$; we have $∀$$g\in G$ ,$g^2\in H$. I suppose that $A_4$ has such this subgroup, named $H$, of order 6. Then for any $\sigma\in A_4$; $\sigma^2\in H$. I think maybe the contradiction happens when we enumerate all $\sigma^2$. May I ask if there is another approach for this problem? Thanks.

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+1 Your approach does work. There are a total of 8 3-cycles in $A_4$. If $\sigma$ is any one of them, then $\sigma^2=\sigma^{-1}$, so they have 8 distinct squares. Together with the identity element your argument then shows that $H$ must have at least 9 elements, which is an obvious contradiction. – Jyrki Lahtonen May 28 '12 at 5:59
+1 For showing your attempt on the problem. – user38268 May 28 '12 at 6:01

Suppose $A_4$ has a subgroup of order 6. Then that subgroup which we call $H$ must be a normal subgroup because the index of $H$ in $A_4$ is 2. Now we know that for any group $G$, if a subgroup say $K$ of $G$ is normal then it must be a union of conjugacy classes. So in our case, $H$ must be a union of conjugacy classes of $A_4$. What are the conjugacy classes in $A_4$?

You already know that the identity element is always in the conjugacy class of itself. Alternatively you could use the fact that the only groups up to isomorphism of order 6 are the cyclic group of order 6 or $S_3$.

Suppose we now view $A_4$ as a subgroup of $S_4$. If it has a subgroup order $6$ sitting inside of it, it cannot be the cyclic group of order 6 because $A_4$ has no element of order $6$. It is also impossible for $S_3$ to be contained inside of $A_4$ because $S_3$ has odd cycles. It follows that $A_4$ has no subgroup of order 6.

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I know that number of conjugacy classes is 4. As GAP shows: [ ()^G, (1,2)(3,4)^G, (1,2,3)^G, (1,2,4)^G ] – Babak S. May 28 '12 at 5:48
1,3,4,4 respectively. Thanks I got the point. Thanks – Babak S. May 28 '12 at 5:54
@BabakSorouh Now the size 1 conjugacy class corresponds to the identity. It follows that we must now have some union of conjugacy class where the total size of the union is 5. You already know that conjugacy classes are disjoint. Therefore the size of the union will be the sum of the sizes of the individual conjugacy classes. In other words, you are trying to write 5 as a sum of the numbers $3$ and $4$. Is this possible? – user38268 May 28 '12 at 5:55
It is certainly impossible. Thanks – Babak S. May 28 '12 at 6:00
@BabakSorouh Right. Sorry I did not answer about your approach, I thought you were asking for another approach which is what I gave above. – user38268 May 28 '12 at 6:01

Suppose there exists a subgroup $H$ of order $6$, so $[A_4:H]=2$. Now there are $8$ $3$-cycles in $A_4$, so there exists a $3$-cycle $g\notin H$. Then consider the cosets $H, gH, g^2H$. So two must coincide. Since $H\neq gH$, $g^2H$ must equal one of the others, but either case implies $g\in H$, a contradiction.

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Wait, why can that $g^2$ belong to $H$ show that $g$ belongs to $H$? – awllower May 28 '12 at 11:21
If $g^2 \in H$, then $(g^2)^2 = g^4 = g \in H$. – Dustan Levenstein May 28 '12 at 12:09

Your approach and many more can be found in this article, where $11$ different proofs are given.

Michael Brennan. Des Machale. Variations on a Theme: $A_4$ Definitely Has no Subgroup of Order Six!, Mathematics Magazine, Vol. $73$, No. $1$ (2000) JSTOR

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Well, I am wondering if I could use the characters... If so, then here as follows:
As an exercise, show that, if G quotient its center is abelian, and if its commutator subgroup is of prime order p, then every non-linear irreducible character must be of degree n such that n²=|G:Z(G)|.
However, as H is of index 2, it is normal and its quotient group is abelian, that is, the commutator subgroup G' of G is either of prime order or equal to H. If the former case, then there is a non-linear character whose degree² is = 2, not possible. Hence G'=H. But then G/G' is cyclic, so G is abelian, which is not. Therefore, no such H can exist. Maybe there is some gap, as this argument shows that no group of order 12 can have subgroups of order=6...
Inform me if the gap is detected, thanks.

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I found one gap: H is not necessarily abelian!! bUT Then H is isomorphic to $S_3$, which we might show to be impossible within A4. – awllower May 28 '12 at 10:50
So this actually proves that, if a group G of order=12 has an abelian subgroup of order=6, then G must be abelian. Thus the question reduces to showing that any subgroup of A4 of order=6 is abelian, which might serve as an exercise. – awllower May 28 '12 at 11:17