I'm reading this proof from this article and I don't see why one argument works. In the Lemma, $n_4(G)$ is the number of elements of order $4$ in $G$.
In the first step of the induction we have a group $G$ with $|G|=16$ and $exp(G)=4$, three maximal subgroups $A,B,C \leq G$, s.t. $A \cap B \cap C =H$ with $|H|=4$.
Now the author says, that if $C_G(H)=G$, then $G$ is abelian.
I don't see, why this is true and I think I found a counterexample:
I looked up $SmallGroup(16,3)=C_2^2 \rtimes C_4$ in which exist three maximal subgroups with intersection $H=Z(G) \cong C_2^2$. Clearly $C_G(H)=G$, but $G$ is not abelian. What am I missing?
Edit:
I think the next argument is wrong aswell. If $C_G(H)\not = G$, then $C_G(H)=A$ is indeed abelian, but its not true, that $B\cong C$. See $SmallGroup(16,13)$ and chose $Q_8$ as a maximal subgroup. It will always lead to $(A,B,C)=(C_4 \times C_2,Q_8,D_8)$ (up to ordering). Is there any quick way to see that all groups of order $16$ contain a number of elements of order 4, that is divisible by $4$?
Best Answer
I think the parts of the proof which you rightly question aren't actually necessary.
We have $n_4(G)\equiv n_4(A)+n_4(B)+n_4(C)\pmod 4$. Since $C_G(H)\ne H$ we can suppose $A$ is abelian and therefore $n_4(A)\equiv 0\pmod 4$.
If both $B$ and $C$ are non-abelian
Then $n_4(B)\equiv n_4(C)\equiv 2\pmod 4$ and we are finished.
If, say, $B$ is abelian
Then $G=<A,B>\leq C(H)$ and so $C$ is also abelian. Then $n_4(B)\equiv n_4(C)\equiv 0\pmod 4$ and we are finished.