Here is my attempt at proving $$\mathbb{Z}_4 \times \mathbb{Z}_2= \langle x, y \mid x^4=y^2=1, x^y=x \rangle$$

Let $F$ be a free group upon $\{x,y\}$. Define a homomorphism $\theta: F \rightarrow \mathbb{Z}_4 \times \mathbb{Z}_2$ as follows: $$x\theta = (1,1),~ y\theta = (2,0)$$ Then by Von Dyck's theorem, there exists an epimorphism $$\phi: \langle x, y \mid x^4=y^2=1, x^y=x \rangle \rightarrow \mathbb{Z}_4 \times \mathbb{Z}_2$$

Now I just need to show both groups have the same cardinality in order to conclude $\phi$ is an isomorphism.

How can I do this?

  • 2
    $\begingroup$ What does $x^y$ mean in this context? $\endgroup$ – Alex Provost May 7 '17 at 14:40
  • 1
    $\begingroup$ @AlexProvost$ x^y=y^{-1}xy$ $\endgroup$ – mrs May 7 '17 at 14:43
  • $\begingroup$ @ResidentDementor Interesting. I suppose this makes sense in light of $(x^y)^z = x^{yz}$. $\endgroup$ – Alex Provost May 7 '17 at 14:48

Since you have the relations $x^y = x$ and $y^2 = 1$, you know that $x$ and $y$ commute, and so everything in the presented group may be brought to the form $\{x^i y^j \mid 1 \leq i \leq 4, 1 \leq j \leq 2 \}$, and so the cardinality of the presented group is at most 8. Since $\phi$ is a surjection, you know the cardinality is at least $|\mathbb{Z}_4 \times \mathbb{Z}_2| = 8$.


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