# On the Normality of Subgroups in Finite Groups

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 $$G$$.

Thank you.

Consider cases where $$G=N\times M_1$$ and $$M. Then we trivially have $$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.]

• 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?
– Paul
Feb 16 at 21:21
• 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. Feb 16 at 21:24
• 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. Feb 16 at 21:40
• Thanks for your valuable remark. Yes, the original question was meant to specify that $MN=G$ and I have edited it.
– Paul
Feb 16 at 21:48
• (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 $G=MN$.) Feb 16 at 22:53

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 $$2$$.