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Let $G_1$ and $G_2$ be finite solvable groups and $M$ be a maximal subgroup of $G=G_1\times G_2$ show that one of the following holds: $$ G_1\times\{e\}\leq M,\ \{e\}\times G_2\leq M,\ M \unlhd G.$$

We know that $G$ is solvable then since $M\leq G$, $M$ is also solvable. And $[G:M]=p^k$ for some prime p.(Because maximal subgroup of a solvable group has prime power index.)

I couldn't make them useful for this question.

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    $\begingroup$ Assume that neither $G_1$ nor $G_2$ is contained in $M$. So $M$ projects onto both $G_1$ and $G_2$, and hence $M \cap G_1$ and $M \cap G_2$ are normal subgroups of $G$. We can effectively factor out $(M \cap G_1) \times (M \cap G_2)$, and thereby assume that they are both trivial. So now $M$ is a diagonal subgroup of $G_1 \times G_2$. Can you take it from there? (There is a general result that for any group $X$, a diagonal subgroup of $X \times X$ is maximal if and only if $X$ is simple.) $\endgroup$ – Derek Holt Jan 7 '14 at 11:35
  • $\begingroup$ @DerekHolt what do you mean with $M$ projects onto $G_1$ and $G_2$? $\endgroup$ – ortmat Jan 7 '14 at 17:11
  • $\begingroup$ $\{ g_1 \in G_1 \mid \exists g_2: (g_1,g_2) \in M \} = G_1$ and $\{g_2 \in G_2 \mid \exists g_1: (g_1,g_2) \in M \}=G_2$. $\endgroup$ – Derek Holt Jan 7 '14 at 17:14

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