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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.

Servaes
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  • 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. – Berci Dec 02 '18 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. –  Dec 02 '18 at 22:01
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    A group is abelian if all pairs of elements commute. A group is not abelian if some pair of elements does not commute. – Servaes Dec 02 '18 at 22:02

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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 $$(1\ 2\ 3)(1\ 3\ 2)=(1\ 3\ 2)(1\ 2\ 3).$$

Servaes
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