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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.2012-05-28

4 Answers 4

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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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    Ok, I have two questions: (1) What does mean index of group ? (2) Why : "Obviously, for any group G and a subgroup H of it with index 2" ?2014-09-02
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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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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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    If $g^2 \in H$, then $(g^2)^2 = g^4 = g \in H$.2012-05-28
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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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    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.2012-05-28