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Can anyone provide with an explanation of why the group $A_4$, which is the group formed by the set of even permutations of $S_4$ under the operation of composition of functions, can not have an order of $6$? I know Lagrange's Theorem tells us that the orders of possible subgroups of $A_4$ are $1,2,3,4,6,12$, and I can find a subgroup a subgroup of all of the orders listed except $6$, and I'm pretty sure there is not one, but cannot come up with a solid explanation as to why this cannot happen?

Thanks for the help!

Joe
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A subgroup of index $2$ is always normal. All subgroups of order $3$ in $A_{4}$ are conjugate, and furthermore, $A_{4}$ is generated by its $3$-cycles. Hence no proper normal subgroup of $A_{4}$ can contain a $3$-cycle. But any subgroup of order $6$ contains an element of order $3$, a contradiction.

Geoff Robinson
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As $A_4$ has order$12$, a subgroup $H$ of order $6$ would be normal, and $A_4/H$ would be a group of order $2$. Hence if $\sigma\in A_4$, $\sigma^2\in H$; in particular the squares of $3$-cycles are in $H$.

However, the set of squares of $3$-cycles is just the same as the set of $3$-cycles. Thus all $3$-cycles belong to $H$. Unfortunately, the $3$-cycles are $8$.

Bernard
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