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I've read here that such a group has the form $Q_8 \times B \times D$, so my question really is why can't $G$ be direct product like this? Also, I'd like to know what happens if $|G|=200$?

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Mark
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2 Answers2

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$G$ has normal subgroups of order $4$ and $25$ respectively say $H,K$.

Then $G$ is direct product of $H$ and $K$ since $\gcd(o(H),o(K))=1$

Since every group of order $p^2$ is abelian so are $H,K$ and hence so is $G$

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If all subgroups are normal, the Sylow subgroups are unique and $G=HK$ for a $2$-Sylow $H$ and a $5$-Sylow $K$ which centralize each other. They are both of square prime order, so they are both Abelian, and thus $G$ is Abelian.

If $|G|=200$ it is no longer certain the Sylows are Abelian. You could have a $2$-Sylow that looks like the quaternion group.

rschwieb
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  • Is it possible that when the group $G$ of order $200$ is nilpotent then every subgroup of $G$ is normal? – Mark Dec 14 '16 at 19:33
  • @user264885 Isn't $C_8\oplus C_{25}$ a counterexample? I'm not 100% sure.. I was just looking for groups that weren't of the Hamiltonian form you mentioned. The $2$-sylow doesn't seem to fit the description. – rschwieb Dec 14 '16 at 19:54
  • Thanks a lot for your help, I will check if that is true. – Mark Dec 14 '16 at 19:59
  • @user264885 Sorry no, I didn't realize Hamiltonian groups are defined to be nonabelian. My example is not a counterexample. – rschwieb Dec 14 '16 at 20:10
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    @user264885 There is another nonabelian group of order 8 it looks like. I think if you use that with $C_{25}$, you have a counterexample. – rschwieb Dec 14 '16 at 20:14