# How to prove that semidirect product of $Z_{13}$ and $Z_3$ is non Abelian for a non-trivial homomorphism

The semidirect product of $$Z_{13}$$ and $$Z_3$$ is given here Finding presentation of group of order 39 as $$\{x,y | x^{13} = y^3 = 1, yxy^{-1} = x^3\}$$. I understand how this is arrived at but to show that this group is non abelian, if I take $$(x_1,y_1)$$ and $$(x_2,y_2)$$ and look at
$$(x_1,y_1).(x_2,y_2) = (x_1 \phi_{y_1} (x_2),y_1y_2) = (x_1y_1x_2y_1^{-1},y_1y_2) = (x_1x_2^3,y_1y_2)$$
$$(x_2,y_2).(x_1,y_1) = (x_2 \phi_{y_2} (x_1),y_2y_1) = (x_2y_2x_1y_2^{-1},y_2y_1) = (x_2x_1^3,y_2y_1)$$
$$y_1y_2 = y_2y_1$$ because $$y_1,y_2 \in Z_3$$
$$x_1x_2^{3}\neq x_2x_1^3 \iff x_2x_1x_1^2 \neq x_1x_2x_2^2 \iff x_1\neq x_2 \because x_1x_2 = x_2x_1 \because x_1,x_2 \in Z_{13}$$
But this means $$(x_1,y_1)$$ and $$(x_1,y_2)$$ commute? I am doing something simple wrong here. Any help is appreciated. Thanks.

• No, it's all correct. If $x_1\ne x_2$, then $(x_1, y_1)$ and $(x_2,y_2)$ don't commute, hence the group is not commutative. Commented Dec 2, 2018 at 21:54
• @Berci But if $x_1 = x_2$ and $y_1 \neq y_2$, it should still not commute, right? But it does above.
– user621937
Commented Dec 2, 2018 at 22:01
• A group is abelian if all pairs of elements commute. A group is not abelian if some pair of elements does not commute. Commented Dec 2, 2018 at 22:02

Your equivalences show that if $$x_1\neq x_2$$ then $$(x_1,y_1)$$ and $$(x_2,y_2)$$ do not commute, whatever $$y_1,y_2\in\Bbb{Z}_3$$ are. So the semidirect product contains elements that do not commute. This means the semidirect product is not abelian.
Note that a group that is not abelian may still contain elements that do commute with eachother. For example, any element in any group commutes with all its own powers, because $$g\cdot g^k=g^{k+1}=g^k\cdot g$$. A concrete example is $$S_3$$; here we have $$(1\ 2)(1\ 3)\neq(1\ 3)(1\ 2),$$ so this group is not abelian. But still $$(1\ 2\ 3)(1\ 3\ 2)=(1\ 3\ 2)(1\ 2\ 3).$$