0
$\begingroup$

Consider a finite group $G$ with subgroups $N$ and $M$, where $N$ is normal in $G=MN$, $N \cap M = 1$, and $[M, N] \leq \Phi(N)$, where $[M, N]$ denotes the commutator subgroup of $M$ and $N$, and $\Phi(N)$ represents the Frattini subgroup of $N$. The question arises: is $M$ necessarily normal in $G$?

To explore this question further, let's consider the symmetric group $S_3$, where $N = A_3$ (the alternating subgroup of order $3$) and $M = \langle (1,2) \rangle$ (the subgroup generated by the permutation $(1,2)$).

It can be observed that $[N, M] = N$, which is not contained in $\Phi(N) = 1$. In this example failure to have the condition $[M, N] \leq \Phi(N)$ leads to conclude that $M$ is not normal in $G$.

Please provide your insight or suggestions.
Thank you.

$\endgroup$

2 Answers 2

4
$\begingroup$

Consider cases where $G=N\times M_1$ and $M<M_1$. Then we trivially have $N\triangleleft G$ and $N\cap M=1$, and things in $M,N$ commute with one another so $[M,N]\leq\textrm{anything}$. But if $M$ isn't a normal subgroup of $M_1$ then it isn't a normal subgroup of $G$ either.

[When the above was written, the question did not specify that $G=MN$. It does not provide a counterexample to the question as it now is, with that added condition.]

$\endgroup$
5
  • $\begingroup$ Thanks Gareth for your response. However, I can not see how is your example will fit the problem as stated. In the suggested construction where $G = N \times M_1$ and $M < M_1$, we are claiming that the complement for the normal subgroup (in this case, $M_1$) will be normal in $G$ and that is still true in your construction, or am I missing some thing? $\endgroup$
    – Paul
    Feb 16 at 21:21
  • $\begingroup$ I'm not sure I understand the question. (I'm not a group theorist so it's very possible that I'm just confused.) What you originally asked was whether the conditions you gave imply that $M$ is normal in $G$, not whether they imply that "the complement for the normal subgroup will be normal in $G$". Is it possible that the original question was meant to specify that $MN=G$? Unless I'm missing something, it doesn't say that. $\endgroup$ Feb 16 at 21:24
  • $\begingroup$ But in the problem statement you didn't write $G=NM$. You just edited that in four minutes ago. I agree that my counterexample is not a counterexample to what the question is asking now. $\endgroup$ Feb 16 at 21:40
  • $\begingroup$ Thanks for your valuable remark. Yes, the original question was meant to specify that $MN=G$ and I have edited it. $\endgroup$
    – Paul
    Feb 16 at 21:48
  • $\begingroup$ (In case it isn't obvious, my comment beginning "But in the problem statement" was a reply to something Paul wrote, since deleted, that implied that the problem statement in the question said $G=MN$.) $\endgroup$ Feb 16 at 22:53
0
$\begingroup$

No $M$ is not necessarily normal in $G$. A counterexample is the dihedral group of order $8$, with $N$ cyclic of order $4$ and $M$ a non-normal subgroup of order $2$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .