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Use the following lemma to prove that $A_4$ has no subgroup of order $6$:

Lemma: If $H\le G$ has index $2$, i.e. $[G:H]=2$, then for any $a\in G$ we have $a^2\in H$.

The $12$ elements of $A_4$ are $(1), (12)(34), (13)(24), (14)(23), (123), (132), (124), (142), (134), (143)$, $(234)$, and $(243)$.

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  • $\begingroup$ If we see from the lemma that the index is 2 can we say that since we know |G| = 12 then by Lagrange's theorem |G| = [G:H] * |H| => 12 = 2 * |H| => 6 = |H|, why is this not applicable and A4 has no subgroup of order 6? $\endgroup$ – Arnold Oct 30 '13 at 3:04
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The order of $A_4$ is 12, so if $A_4$ has a subgroup $H$ of order $6$, then $[A_4:H]=2$ (why?). Then it follows from the lemma that $a^2\in H$ for all $a\in A_4$. List the elements of the set $\{a^2\mid a\in A_4\}$, that is, all squares in $A_4$. How many are there?

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HINT: If $H$ were a subgroup of $A_4$ of order $6$, then $H$ would have index $2$ in $A_4$, so the square of every element of $A_4$ would belong to $H$. How many elements are in the set $\{a^2:a\in A_4\}$?

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