# If $O^{2'}(G) / N$ simple for normal $2'$-subgroup, then $G$ has no subgroup of index $2$

Let $G$ be a finite group. A group is called a $2'$-group if its order is odd. Also $O^{2'}(G)$ denotes the smallest normal subgroup such that $G / O^{2'}(G)$ has odd order.

If there exists some normal $2'$-subgroup $N$ of $G$ such that $O^{2'}(G) / N$ is simple, then $G$ could not contain a subgroup of index two.

Is this true, and if so why?

• Why not look at some small groups to see if you can find a counterexample? – Derek Holt Nov 6 '15 at 8:46
• Even $C_2$ is a counterexample. I was wondering whether they might have meant the $O^{2'}(G)/N$ is a nonabelian simple group. Then the claim would be true. (In fact $O^{2'}(G)/N = C_2$ is the only simple group that leads to the statement being wrong.) – Derek Holt Nov 6 '15 at 12:50
• Well, we have $1 \le N < O^{2'}(G) \le G$, where $N$ and $G/O^{2'}(G)$ have odd order, so if $G$ has a subgroup of index $2$ then $O^{2'}(G)/N$ must have one too! – Derek Holt Nov 6 '15 at 13:36
• More generally if a group $G$ has a subnormal series $G \unrhd G_1 \unrhd \cdots \unrhd G_1=1$ and a normal subgroup of prime index $p$, then at least one of the quotients $G_i/G_{i+1}$ must have a normal subgroup of index $p$. – Derek Holt Nov 6 '15 at 18:02
• I don't understand your question about why $N < O^{2'}(G)$. That was part of your assumptions! – Derek Holt Nov 7 '15 at 13:45