# Any group of odd square free order is cyclic

So this problem is actually mistaken, the condition should really be that a group of order $n$ with $\gcd(n,\varphi(n))=1$, not just being odd square free. Since there exists a group of order 21 which is not abelian, thanks to @NickyHekster. Sorry for the fallacy conclusion. So I posted a new one with the complete problem here: Any group of order $n$ satisfying $\gcd (n, \varphi(n)) =1$ is cyclic

• Where do you get stuck? Did you manage to find a section of your epimorphism $G \to H$? Dec 1, 2013 at 10:09
• @user It was asked by the problem... Dec 1, 2013 at 10:23
• @PatrickDaSilva I get stuck on part (b), to show G is isomorphic to $N \times H$ Dec 1, 2013 at 10:24
• Sorry if I'm misunderstanding, but is $\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}$ not a counterexample to this? Dec 1, 2013 at 10:25
• This can't be correct. Note that $|\text{Aut}(\mathbb{Z}_{11})|=10$, and so there is a non-trivial homomorphism $\varphi:\mathbb{Z}_5\to \text{Aut}(\mathbb{Z}_{11})$ and so there is a non-abelian group of order $55$ given by $\mathbb{Z}_{11}\rtimes_\varphi\mathbb{Z}_5$. Dec 1, 2013 at 10:39

It is not true, there is a non-cyclic group of order 21.

• Which one? I can only think of $\mathbb Z / 21 \mathbb Z$. If you want to convince us, give us an example! Dec 1, 2013 at 10:32
• Take a semi direct product of $C_7$ and $C_3$ Dec 1, 2013 at 10:40
• Hm, strange ; one needs a counterexample with $p < q$ and $q \equiv 1 \pmod p$, so that the first nontrivial example comes at quite a large integer (I mean, $21$ is not that small, even though it is not the monster group.) Dec 1, 2013 at 10:45
• @NickyHekster, in your last comment you must have meant "The smallest simple non-abelian group..." Dec 1, 2013 at 12:35

To find the section, use the Sylow theorems.

You have found $\pi : G \to H \simeq \mathbb Z / p \mathbb Z$, an epimorphism. Now in $G$, since $|G| = p_1\cdots p_s$, without loss of generality assume $|H| = p_s$. By Lagrange's theorem there exists $x \in G$ with $K \overset{def}=\langle x \rangle \simeq \mathbb Z / p \mathbb Z$.

Using the Sylow theorems, $K$ is a Sylow $p$-subgroup since $|G|$ is squarefree. We wish to show that $K \le G$ is the unique subgroup of $G$ of order $p$. Call the number of such subgroups $n_p$. Since $n_p$ divides $|G|/|K| = p_s$ and $n_p = 1 \pmod{p_s}$, $n_p = 1$. That is, $K$ is unique. Any isomorphism $H \to K$ gives rise to a section $H \hookrightarrow G$. Using the expansion of my suggestion by Giorgio Mossa, you're done.

Added : For those who worry that this argument is wrong, it is not. This is because the problem does not lie here. Using this section, one gets a normal subgroup $N \trianglelefteq G$ and a subgroup $H \le G$ such that $NH = G$ and $N \cap H = \{e\}$, i.e. one gets a semi-direct product. Therefore the most general result one can get is a decomposition $$G \simeq (\cdots(( H_1 \rtimes H_2 ) \rtimes H_3 ) \rtimes \dots \rtimes H_s$$ where $|H_i| = p_i$, using induction on $i$ up to $s$ where $|G| = p_1 \cdots p_s$.

Hope that helps,

• @Old John : There is no fallacy in my argument. The problem probably hides somewhere else. Dec 1, 2013 at 10:43
• @Old John : Actually the section does exist. The thing is that this section gives rise to a semidirect product, so the problem is just that one will not be able to prove that this section gives a direct product. (It will give you all these other examples of semi-direct products we keep mentioning.) Dec 1, 2013 at 10:51
• @Old John : Found the problem, explained it above. Dec 1, 2013 at 10:57
• Thanks a lot! But as I edited in the problem, I missed a condition: $\gcd(\varphi(n),n)=1$. If we know this, how do we conclude that $K$ is normal? i.e. is the unique Sylow $p$-subgroup? Dec 2, 2013 at 8:38
• @user112564 : Just try to show that any automorphism $\varphi : H \to \mathrm{Aut}(N)$ has to be trivial. Then all these direct products I explicited above have to be trivial! Using your order condition this should be quite simple. Dec 4, 2013 at 0:03

Here are some hint that can help you in proving the part b).

When you have a surjective homomorphism

$$\pi \colon G \to H$$ and want to show that $G \cong H \times N$ for $N=\ker \pi$ a standard technique is to find a morphisms $i \colon H \to G$ such that $\pi \circ i=1_H$, that's the section which Patrick Da Silva was referring to.

If you find such a map then clearly

• $i$ must be injective, since $\pi\circ i=1_H$ and $\ker i \subseteq \ker \pi\circ i= 0$;
• there's a subgroup $H'=\text{Im } i < G$ isomorphic to $H$, because $i$ is injective;
• the intersection $H' \cap N=(0)$, because $\pi \circ i=1_H$ there cannot be any element in $\text{Im }i$ which is element of $N$

from that you should easily get that $G$ is a semidirect product of $H$ and $N$, to have a direct product you just need to prove that $H'$ is a normal subgroup of $G$ (hint: $H'$ have order $p$ for some prime that divide the order of $G$, and is the maximal power of $p$ dividing it).

The last part is to prove that $N$ is an abelian group, in order to prove that $G$ is abelian (hint: induction).

Hope this helps.

• Thanks for expanding my suggestion. Dec 1, 2013 at 10:28
• @GiorgioMossa Thanks a lot! But as I missed a crucial condition that $\varphi(n)$ is relatively prime to $n$, I think we have to use that to show $H^\prime$ is normal, i.e. to show that $H^\prime$ is the unique Sylow $p$-subgroup. But I still can't come up with that. Dec 2, 2013 at 8:46