# A difficult exercise about the first isomorphism theorem of a group

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 you show that $G/Z(G)$ is not commutative? How may non commutative group are there with 6 elements? – Sheve Apr 30 at 15:23
• 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). – Greg Martin Apr 30 at 16:02
• I tried that but I did not reach the goal – user149240 Apr 30 at 16:22

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?
• It's exactly this: we define homomorpshism $h: G \to S_3$ s.t. kernel of $h$ is $Z(G)$ and $h$ is surjective. – mihaild May 1 at 12:36