Algebra Proof Question

1. Dec 6, 2007

JasonRox

I have questions about this proof. First question is the italic part.

A finite group G with no characteristic subgroups other than G and 1 is either simple or a direct product of isomorphic simple groups.

Proof: Choose a minimal normal subgroup H of G whose order is minimal among all nontrivial normal subgroups. Write H = H_1, and consider all subgroups of G of the form $H_1 x ... x H_n$, where n >= 1, H_i is normal in G and H_i is isomorphic to H. Let M be such a subgroup of largest possible order. We show that M = G by showing that M char G; to see this, it suffices to show that $\alpha(H_i)$ is contained in M for every i and every automorphism $\alpha$ of G. Of course, $\alpha(H_i)= H = H_1$. We show that $\alpha(H_i)$ is normal in G. If a is in G, then $a = \alpha(b)$ for some b in G, and $a\alpha(H_i)a^{-1}$ = $\alpha(bH_{i}b^{-1})$ is contained H_i since it is normal in G. If $\alpha(H_i)$ is not contained in M, then $\alpha(H_i) n M$ (n is intersect) is not contained in $\alpha(H_i)$ and $|\alpha(H_i) n M| < |\alpha(H_i)| = H$. But $\alpha(H_i) n M$ is normal in G, so this implies that the order of that group is 1 because of the minimality of H. The subgroup <M,$\alpha(H_i)$>=M x $\alpha(H_i)$ is a subgroup of the same type as M but of larger order, a contradiction. We conclude that M char G, and so M = G. Finally, H = H_1 must be simple: if N is a nontrivial normal subgroup of H, then N is a normal subgroup of M = H_1 x ... H_n = G, and this contradicts the minimal choice of H.

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Nevermind, I got it.

You can read it for pleasure I guess.

Last edited: Dec 6, 2007
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