# Use a lemma to prove that $A_4$ has no subgroup of order $6$. [duplicate]

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 $$(243)$$.

• 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? – Arnold Oct 30 '13 at 3:04

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?
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 $\{a^2:a\in A_4\}$?