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I was helping a friend of mine studying basic algebra, but I waz stuck in this exercise.

Let $G=\mathbb{Z}_6 \times \mathbb{Z}_6 $ with the operation

$(a, b) *(c, d) =(a+(-1)^b c, b+d) $. Prove that $G/Z(G) $ is isomorphic to $S_3$.

I have proved that G is a group and that the center of G is the set that contains (0,3), (2,3) (4,3), (0,0), (2,0), (4,0).

Now, to prove the isomorphism, I would like to define an homomorphism between $G$ and $S_3$, whose kernels is exactly $Z(G) $, so that for the first theorem of homomorphism of groups they are isomorphic.

But the problem is that I have no idea how to define the homomorphism.

Can anyone help me, please?

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    $\begingroup$ Can you show that $G/Z(G)$ is not commutative? How may non commutative group are there with 6 elements? $\endgroup$ – Sheve Apr 30 at 15:23
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    $\begingroup$ Alternatively, start looking at the orders of the elements of $G/Z(G)$, since you know any isomorphism from $G/Z(G)$ must send an element of order 2 (or 3) to an element of $S_3$ of order 2 (or 3). $\endgroup$ – Greg Martin Apr 30 at 16:02
  • $\begingroup$ I tried that but I did not reach the goal $\endgroup$ – user149240 Apr 30 at 16:22
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Lets start from any element not from center - for example, $(0, 1)$. We have $(0, 1) * (0, 1) = (0, 2)$ and $(0, 2) * (0, 1) = (0, 3) \equiv (0, 0)$. So order of $(0, 1)$ is $3$. Say that $(0, 1)$ goes to $(123)$, and then $(0, 2)$ goes to $(132)$.

Take not yet covered element, for example $(1, 0)$. We have $(1, 0) * (1, 0) = (2, 0) \equiv (0, 0)$. So order of $(1, 0)$ is $2$. Say that $(1, 0)$ goes to $(12)$.

Now $(1, 0) * (0, 1) = (1, 1)$ goes to $(12) \cdot (123)$ = $(23)$. So we found pre-images of $(123)$, $(132)$, $(12)$ and $(23)$. This also defines that elements goes to the rest of $S_3$. Can you complete the definition and prove that it is indeed a homomorphism?

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  • $\begingroup$ Yes of course.I have thought about seeing where an element goes, but I would like to use the first theorem of isomorphism. But thanks for the advice $\endgroup$ – user149240 May 1 at 12:28
  • $\begingroup$ It's exactly this: we define homomorpshism $h: G \to S_3$ s.t. kernel of $h$ is $Z(G)$ and $h$ is surjective. $\endgroup$ – mihaild May 1 at 12:36

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