I have a semi-direct product that I feel must be nonabelian, but my thought process is telling me it is abelian.

I have $G\cong Z_7\rtimes Z_4$ and I have indeed found nontrivial homomorphisms $\phi:Z_4\rightarrow Z_6$ so this guy must be nonabelian correct? Also, if so, then by a proposition in my book (Dummit and Foote) I have that $Z_4$ is not normal in $G$ and so by Sylow's theorem there are 7 groups of order 4 in $G$ and they must all be conjugate to each other. Now I also know from my book that if any subgroup is conjugate to another subgroup then the two subgroups are isomorphic. This means that every Sylow 2-subgroup of $G$ is isomorphic to $Z_4$ and thus all of my subgroups of $G$ are cyclic. Doesn't this mean that $G$ must also be cyclic and therefore abelian?

In general I am trying to classify all groups of order 28 and I am having a hard time showing that the group I have above is distinct from the other groups I have ($Z_{28},Z_2\times Z_{14},$ and $Z_7\rtimes V_4$).

Thanks in advance!

  • $\begingroup$ all subgroups of $S_3$ are cyclic. $\endgroup$
    – clark
    Oct 27 '13 at 3:00
  • $\begingroup$ Thank you! I should have tried thinking of counter examples. $\endgroup$ Oct 27 '13 at 3:08

First of all, yes, if you have a non-trivial homomorphism $\mathbb{Z}_4\to\mathbb{Z}_6$, then the corresponding semi-direct product is going to be non-abelian. Your mistake is in assuming that all subgroups of $G$ are cyclic. Yes, all $p$-subgroups of $G$ are cyclic, but, for example, if $x$ is the generator of $\mathbb{Z}_7$ and $y$ is the generator of $\mathbb{Z}_4$, then the group generated by $x$ and $y^2$ is not cyclic.

Note As a user pointed out in a comment above, the fact that all proper subgroups of a given group are cyclic does not mean that the group is cyclic, or even abelian! So first mistake is assuming this, and the second mistake is thinking that in the group above all subgroups are cyclic (even though this doesn't influence the proof of non-commutativity)!

So it's enough just to say that the homomorphism you found isn't trivial. This is because this homomorphism is going to be the way you conjugate certain elements in $G$. If the homomorphism is non-trivial, then there is going to be non-trivial conjugation in $G$, and so it can't be abelian.

  • $\begingroup$ Ahhh, shoot, that's so elementary. Thanks! :) $\endgroup$ Oct 27 '13 at 3:07
  • $\begingroup$ No problem! Good question! $\endgroup$
    – rfauffar
    Oct 27 '13 at 3:10

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