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We know that when a group $G$ has order $2^k m$, where $m$ is an odd integer, $G$ should have a normal subgroup with order $m$ from here. When $k=1$, this implies the index of the normal subgroup is $2$. However, when $k=2$, $m$ is a prime, can we also find a normal subgroup which has index $2$?

That is to say, when $|G|=4p$, where $p$ is a prime congruent to $1$ or $3$ modulo $4$, does a normal subgroup of index $2$ always exists? I don't know how to construct the contourexample, any idea or solutions are welcomed, thanks!

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The alternating group of order 12 is a counterexample to the first sentence, and an answer to your question, I think. –  Jack Schmidt May 24 '13 at 4:25

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up vote 4 down vote accepted

If a group $G$ has a normal subgroup $N$ of index $2^k$, then $G/N$ is a group of order $2^k$ and so has normal subgroups of index $2^i$ for all $i=0,1,\ldots,k$, and in particular of index 2. The lattice isomorphism theorem is that the subgroups of $G/N$ are exactly the $H/N$ for subgroups $H$ containing $N$, and that $H/N$ is normal in $G/N$ if and only if $H$ is normal in $G$.

If $G$ is a finite group with a cyclic Sylow 2-subgroup, then $G$ has a normal 2-complement, a subgroup whose index is equal to the order of the Sylow 2-subgroup. By the previous, $G$ has normal subgroups of index $2^i$ for all $2^i$ dividing the order of the Sylow 2-subgroup. As long as $G$ has even order, it must have a normal subgroup of index 2.

If the Sylow 2-subgroup is not cyclic, then $G$ need not have any normal subgroups of index 2, as the alternating group on four points shows.

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