Characteristic groups

**Prometheus** · December 24th 2008, 06:46 AM

I got stuck with two questions dealing with characteristic groups

the first - G is a finite group characteristically simple and N is a minimal normal sub group in G (contains no proper normal subgroups of G). if $H=N_1 \times N_2 \times \cdots \times N_k$ is a direct sum of isomorphic copies of $N=N_1$ where k is maximal, then it is normal in G.

I can show that N is characteristically simple, and also all its isomorphic copies. I tried to show that if $gN_i g^{-1} \cap N_j \neq \{ e \}$ then $gN_i g^-1 \cap N_j = N_j$ , and show that either there is another copy of N that I can join to the sum, or H is normal.

The problem is I can't see how I can use the fact that G is characteristically simple. I know that Z(G) and the commutator subgroups are characteristic in G but couldn't find what to do with it.

The second question is to show that if N is normal in G then $\Phi(N)\subseteq\Phi(G)$ where $\Phi(A)$ is the intersection of all maximal groups in A. I thougt to use $\Phi(G)\cap N$ and show that there is an intersection of some maximal groups in N that gives it. another thing is if H is maximal in G, and $\Phi(N)$ is not contained in H then maybe $H\Phi(G)$ will give a contradictory to the maximality of H (which is well defined bacuase $\Phi(G)$ char N normal in G).

any insights?

**NonCommAlg** · December 24th 2008, 02:18 PM

Originally Posted by Prometheus

the first - G is a finite group characteristically simple and N is a minimal normal sub group in G (contains no proper normal subgroups of G). if $H=N_1 \times N_2 \times \cdots \times N_k$ is a direct sum of isomorphic copies of $N=N_1$ where k is maximal, then it is normal in G.

this is a strange problem! the reason is that i think $H=G$ if $G$ is characteristically simple. that's because $\alpha(N_1)$ is a copy of $N_1$ for any $\alpha \in \text{Aut}(G).$ why? thus $H_0=\prod_{\alpha \in \text{Aut}(G)} \alpha(N_1) \leq H.$

but $H_0$ is a non-trivial characterstic subgroup of $G.$ thus $H_0=G,$ because $G$ is characteristically simple.

The second question is to show that if N is normal in G then $\Phi(N)\subseteq\Phi(G)$ where $\Phi(A)$ is the intersection of all maximal groups in A.

do you know this fact that the Frattini subgroup, $\Phi(G),$ is the set of all the non-generators of G? if you know this, then your problem is not that hard. first note that if G has no maximal subgroup

(this might happen if G is not finite), then by definition $\Phi(G)=G,$ and there's nothing to prove. otherwise, if $\Phi(N) \subseteq M,$ for any maximal subgroup M of G, then we're done. so suppose there

exists a maximal subgroup M of G such that $\Phi(N) \nsubseteq M.$ now since $\Phi(N)$ is characterstic in N and N is normal in G, $\Phi(N)$ is normal in G. thus $\Phi(N)M$ is a subgroup of G properly containing M.

thus $\Phi(N)M=G,$ by maximality of M. therefore $(M \cap N)\Phi(N)=N.$ why? that means $<\{\Phi(N), M \cap N \} >=N$ and hence $M \cap N=N,$ because the elements of $\Phi(N)$ are non-generators.

so we've proved that $N =M \cap N \subseteq M,$ which gives us $\Phi(N) \subseteq N \subseteq M.$ contradiction!

**Prometheus** · December 26th 2008, 12:22 AM

in the first question, I meant that k is maximal in finding k-1 isomorphic copies of $N=N_1$ so that $H=N_1 \times N_2 \times \cdots \times N_k$ is a subgroup of G. not all the copies of N are in there. (or maybe they are, but this needs a proof).

any way, thanks for the second question

**NonCommAlg** · December 26th 2008, 12:43 AM

Originally Posted by Prometheus

in the first question, I meant that k is maximal in finding k-1 isomorphic copies of $N=N_1$ so that $H=N_1 \times N_2 \times \cdots \times N_k$ is a subgroup of G. not all the copies of N are in there. (or maybe they are, but this needs a proof).

any way, thanks for the second question

well, the product of finitely many normal subgroups is obviously normal. for that we don't need so many conditions that are givem in your problem! that's why i think something is missing here!

**Prometheus** · December 26th 2008, 12:56 AM

the isomorphism is between $N$ and $N_i$ . it is not an automorphism $\alpha$ of G, and then taking $\alpha(N) = N_i$ .

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