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I've been pounding my head on my desk attempting to figure out how to start this proof:

If we have a group $G$, a normal subgroup $N\trianglelefteq G$, $G/N \cong \mathbb{Z}/a\mathbb{Z}$, and $N \cong \mathbb{Z}/b\mathbb{Z}$, where $b<a$, and $(b,a) = 1$, how can we show that $G$ is an abelian group?

I'm stuck on how to start this proof. I know the fact that $b$ and $a$ are relatively prime will have something to do with the proof. Hints as to how to approach this are greatly appreciated (but please do not post the entire proof, or at the bare minimum please hide it in spoiler tags).

Shaun
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blackbrandt
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1 Answers1

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I don't think this statement is true.

Consider the dicyclic group of order $12$: this is the nontrivial semidirect product of $\mathbb{Z}/3\mathbb{Z}$ by $\mathbb{Z}/4\mathbb{Z}$ (so $b=3$, $a=4$; thus $b\lt a$ and $\gcd(a,b)=1$), with the nontrivial action of $\mathbb{Z}/4\mathbb{Z}$ on $\mathbb{Z}/3\mathbb{Z}$.

Explicitly: denote the generator of $\mathbb{Z}/3\mathbb{Z}$ by $x$, and the generator of $\mathbb{Z}/4\mathbb{Z}$ by $y$. The group $$\langle x\mid x^3=1\rangle \rtimes \langle y\mid y^4=1\rangle$$ has underlying set $\langle x\rangle \times \langle y\rangle$. The action is given by ${}^yx = x^{-1}$. Thus, the elements have the form $x^ry^s$ with $0\leq r\lt 3$, $0\leq s\lt 4$, and product given by $$(x^ry^s)(x^{\rho}y^{\sigma}) = x^{r+(-1)^s\rho}y^{s+\sigma}.$$ The subgroup generated by $x$ is cyclic of order $3$ and normal. The quotient is generated by $y$ and thus cyclic, of order $4$ since $G$ has order $12$ and $\langle x\rangle$ has order $3$. And the group is nonabelian, since $x^{-1}\neq x$, yet $yx = x^{-1}y$.

In the notation of the problem: $G=C_3\rtimes C_4$. $N=\langle x\rangle\cong\mathbb{Z}/3\mathbb{Z}$; $G/N\cong \mathbb{Z}/4\mathbb{Z}$. We have $3\lt 4$ and $\gcd(3,4)=1$.

More on groups of order $12$; groupswiki is not necessarily always reliable, but this is simple enough that it is unlikely to contain errors. Here's Wikipedia. Note that Wikipedia is realizing the group as an extension of $C_6$ by $C_2$; but the subgroup of order $3$ in $C_6$ is characteristic, hence invariant under the action. If you take the presentation $$\langle a,x\mid a^{6}=1, x^2=a^3, xax^{-1}=a^{-1}\rangle$$ then the subgroup generated by $a^2$ is normal of order $3$, and the quotient is generated by $x$ (using $x^2=a^3$) and thus is cyclic of order $4$.

More generally, if $N$ is nontrivial cyclic of odd order $2n+1$, then inversion is an automorphism of order $2$. If we take any cyclic group of order $2^k$, this group can be made to act on $N$ by inversion, defining a nontrivial semidirect product $C_{2n+1}\rtimes C_{2^k}$. This is nonabelian because the action is nontrivial and $2n+1\gt 1$; it has a normal subgroup of order $2n+1$ and the quotient is cyclic of order $2^k$, with $\gcd(2n+1,2^k)=1$. Make $k$ sufficiently large so that $2n+1\lt 2^k$ and you have other similar examples.

Arturo Magidin
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  • Interesting... so the generalization I tried to make did not work. I was trying to generalize the problem of $N\cong Z/2Z$, and $G/N = Z/5Z$. – blackbrandt Aug 06 '21 at 03:59
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    @blackbrandt: If you were attempting a statement that you were conjecturing, it would really be best to tell people that it was your conjecture. The way you posted, it looks more like an exercise/theorem you encountered and you are trying to prove it. Which was it? – Arturo Magidin Aug 06 '21 at 04:00
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    @blackbrandt: Since the automorphism group of $\mathbb{Z}/2\mathbb{Z}$ is trivial, the action of $G/N$ on $N$ is trivial. Normal subgroups of order $2$ are always central, so everything commutes with all their elements. If $G/N$ is cyclic, then a pullback of the generator together with the nontrivial element of $N$ generates $G$, and the two commute, so $G$ is abelian (though not necessarily cyclic; however, in that case $G$ will in fact be cyclic). – Arturo Magidin Aug 06 '21 at 04:03
  • I'm realizing that in hindsight... I was hoping that the generalization would help me understand the problem more but I didn't realize my generalization was not accurate. Sorry for the misunderstanding! – blackbrandt Aug 06 '21 at 04:06
  • @blackbrandt: If you further assume that $G/N$ has odd order (when $N$ is cyclic of order $2$), then $G$ will necessarily be cyclic. It is abelian even if $G/N$ has even order, but in that case it need not be cyclic; e.g., $C_4$. – Arturo Magidin Aug 06 '21 at 04:15