Characteristic groups

• Dec 24th 2008, 06:46 AM
Prometheus
Characteristic groups
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 $\displaystyle H=N_1 \times N_2 \times \cdots \times N_k$ is a direct sum of isomorphic copies of $\displaystyle N=N_1$ 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 $\displaystyle gN_i g^{-1} \cap N_j \neq \{ e \}$ then $\displaystyle gN_i g^-1 \cap N_j = N_j$, 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 $\displaystyle \Phi(N)\subseteq\Phi(G)$ where $\displaystyle \Phi(A)$ is the intersection of all maximal groups in A. I thougt to use $\displaystyle \Phi(G)\cap N$ 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 $\displaystyle \Phi(N)$ is not contained in H then maybe $\displaystyle H\Phi(G)$ will give a contradictory to the maximality of H (which is well defined bacuase $\displaystyle \Phi(G)$ char N normal in G).

any insights?
• Dec 24th 2008, 02:18 PM
NonCommAlg
Quote:

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 $\displaystyle H=N_1 \times N_2 \times \cdots \times N_k$ is a direct sum of isomorphic copies of $\displaystyle N=N_1$ where k is maximal, then it is normal in G.

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

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

Quote:

The second question is to show that if N is normal in G then $\displaystyle \Phi(N)\subseteq\Phi(G)$ where $\displaystyle \Phi(A)$ is the intersection of all maximal groups in A.
do you know this fact that the Frattini subgroup, $\displaystyle \Phi(G),$ 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 $\displaystyle \Phi(G)=G,$ and there's nothing to prove. otherwise, if $\displaystyle \Phi(N) \subseteq M,$ for any maximal subgroup M of G, then we're done. so suppose there

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

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

so we've proved that $\displaystyle N =M \cap N \subseteq M,$ which gives us $\displaystyle \Phi(N) \subseteq N \subseteq M.$ contradiction!
• Dec 26th 2008, 12:22 AM
Prometheus
in the first question, I meant that k is maximal in finding k-1 isomorphic copies of $\displaystyle N=N_1$ so that $\displaystyle H=N_1 \times N_2 \times \cdots \times N_k$ 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
• Dec 26th 2008, 12:43 AM
NonCommAlg
Quote:

Originally Posted by Prometheus
in the first question, I meant that k is maximal in finding k-1 isomorphic copies of $\displaystyle N=N_1$ so that $\displaystyle H=N_1 \times N_2 \times \cdots \times N_k$ 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!
• Dec 26th 2008, 12:56 AM
Prometheus
the isomorphism is between $\displaystyle N$ and $\displaystyle N_i$. it is not an automorphism $\displaystyle \alpha$ of G, and then taking $\displaystyle \alpha(N) = N_i$.