# Groups of order $pqr$

Let $G$ be a group of order $pqr$ (distinct primes) and $\varphi:G\to G$ with $\varphi(G)$ isomorphic to $\mathbb{Z}/pq\mathbb{Z}$.

Does this make $G$ abelian/cyclic? Is the kernel of $\varphi$ a subset of the center of $\mathbb{Z}$?

• I know it is easier to prove this using Syllow theorems. I am working on some abstract algebra for a graduate programme exam next year. I never had formal instruction in abstract algebra. – user115017 Nov 13 '15 at 21:41

It follows that $\ker(\varphi)$ is isomorphic to $\mathbb Z/r\mathbb Z$, and it's automatically normal. The homomorphism $\varphi: G \to \varphi(G) \simeq \mathbb Z/pq\mathbb Z$ selects a complementary subgroup to $\ker(\varphi)$, i.e., we have $$\ker(\varphi) \varphi(G) = G$$ and $$\ker(\varphi) \cap \varphi(G) = \{1\}.$$

(Exercise: prove this. Note that $\ker(\varphi) \varphi(G)$ is a subgroup because $\ker(\varphi)$ is normal.)

These are the building blocks of a semidirect product, so we have

$$G \simeq \mathbb Z/r\mathbb Z \rtimes \mathbb Z/pq\mathbb Z.$$

If you haven't seen semidirect products before, suffice to say there are many examples of this situation where $G$ is not commutative, and $\ker\varphi$ is not central. If you have seen semidirect products, you should be able to construct some examples.

Edit: For an explicit counterexample that doesn't require you know about semidirect products, consider the group $G = \mathbb Z/5\mathbb Z \times S_3$, where $S_3$ is the symmetry group on $3$ letters. This will form a counterexample with $p = 2, q = 5, r=3$. To see this in detail, we use the homomorphism $sgn: S_3 \to \mathbb Z/2\mathbb Z$, given by $sgn(1) = sgn( (123)) = sgn((132))=0$ and $sgn( (12)) = sgn((13))=sgn((23))=1$. This gives us a homomorphism $$id \times sgn: G \to \mathbb Z/5\mathbb Z \times \mathbb Z/2\mathbb Z,$$ $$(\overline x, \sigma) \mapsto (\overline x, sgn(\sigma)).$$ To convert this into the desired homomorphism $\varphi: G \to G$, simply observe that $\mathbb Z/5\mathbb Z \times \mathbb Z/2\mathbb Z \simeq \mathbb Z/10\mathbb Z$ and $S_3$ contains a copy of $\mathbb Z/2\mathbb Z$ (three copies, actually - one generated by $(12)$, one generated by $(13)$, and one generated by $(23)$).

• Not sure how to use this. I haven't seen semidirect products, yet. – user115017 Nov 13 '15 at 21:41
• The answer to your question is no. Since you haven't seen semidirect products, it'll be tricky to construct an explicit example for you, but I'll think about it. (Anyone else who happens to know of a nice construction is invited to share their thoughts here.) There is a counterexample, for instance, with $p=2$, $q=5$, and $r=3$, and also one with $p=2$, $q=3$, and $r=5$. – Dustan Levenstein Nov 13 '15 at 21:50
• More generally, there is a counterexample with the triple $(p, q, r)$ if and only if $\gcd(pq, r-1) \neq 1$. – Dustan Levenstein Nov 13 '15 at 21:52
• @user115017 see my edit. – Dustan Levenstein Nov 13 '15 at 22:21

This is not true. Consider the following group of matrices over $\mathbb{Z}/7\mathbb{Z}$: $$G= \begin{Bmatrix} \begin{bmatrix} a & b\\ 0 & 1 \end{bmatrix}\colon a,b\in \mathbb{Z}/7\mathbb{Z}, a\neq 0 \end{Bmatrix}.$$ Then $|G|=6.7=2.3.7$. Define $\varphi\colon G\rightarrow G$ by $$\begin{bmatrix} a & b\\ 0 & 1 \end{bmatrix} \mapsto \begin{bmatrix} a & 0\\ 0 & 1 \end{bmatrix}.$$ Then $\varphi$ is a homomorphism, with $$\ker\varphi= \begin{Bmatrix} \begin{bmatrix} 1 & b\\ 0 & 1 \end{bmatrix}\colon b\in \mathbb{Z}/7\mathbb{Z} \end{Bmatrix}\cong \mathbb{Z}/7\mathbb{Z}$$ and $$Im(\varphi) = \begin{Bmatrix} \begin{bmatrix} a & b\\ 0 & 1 \end{bmatrix}\colon a\in \mathbb{Z}/7\mathbb{Z}, a\neq 0 \end{Bmatrix}\cong \mathbb{Z}/(2.3)\mathbb{Z}.$$ The element $\begin{bmatrix} 1 & 1\\ 0 & 1 \end{bmatrix}$ is in kernel, but not commutes with $\begin{bmatrix} 2 & 0\\ 0 & 1 \end{bmatrix}$, hence non-central, and $G$ is non-abelian.

• The example may had been described in terms of generaters and relations; but presenting in matrices makes everything transparent. – Groups Nov 14 '15 at 4:30