# Prove that there is an isomorphism between two groups.

I'm solving the following problem.

Let H be a group and $$\tau_1:H\rightarrow G_1, \tau_2:H\rightarrow G_2,\cdots\tau_n:H\rightarrow G_n$$ homomorphims with this property: Whenever $$G$$ is a group and $$g_1:G\rightarrow G_1, g_2:G\rightarrow G_2,\cdots g_n:G\rightarrow G_n$$ are homomorphisms, then there exists a unique homomorphism $${g}^*:G\rightarrow H$$ such that $$\tau_i\circ {g}^*=g_i$$ for every $$i.$$ Prove that $$H\cong G_1\times G_2\times \cdots \times G_n.$$

Actually I solved several exercises regarding projection homomorphism before this, so I first chose $$G = G_1\times G_2\times \cdots \times G_n, g_i=\pi_i$$ for all $$i$$ (usual projection mapping from $$G_1\times G_2\times \cdots \times G_n$$ to $$G_i$$). Then, I tried to show that a function $$f:H\rightarrow G_1\times G_2\times \cdots \times G_n$$ defined as $$f(h) = (\tau_1(h),\tau_2(h),\cdots,\tau_n(h))$$ is an isomorphism.

1. $$f$$ is a homomorphism.

Since $$\tau_1,\tau_2,\cdots,\tau_n$$ are homomorphisms,

$$f(ab)=(\tau_1(ab),\tau_2(ab),\cdots,\tau_n(ab))=(\tau_1(a)\tau_1(b),\tau_2(a)\tau_2(b),\cdots,\tau_n(a)\tau_n(b))=(\tau_1(a),\tau_2(a),\cdots,\tau_n(a))(\tau_1(b),\tau_2(b),\cdots,\tau_n(b))=f(a)f(b).$$

2. $$f$$ is surjective.

For $$(y_1,y_2,\cdots,y_n) \in G_1\times G_2\times \cdots \times G_n,$$

$$f({g}^*((y_1,y_2,\cdots,y_n))) = (\tau_1({g}^*((y_1,y_2,\cdots,y_n))),\tau_2({g}^*((y_1,y_2,\cdots,y_n))),\cdots,\tau_n({g}^*((y_1,y_2,\cdots,y_n)))) = (\pi_1((y_1,y_2,\cdots,y_n)),\pi_2((y_1,y_2,\cdots,y_n)),\cdots,\pi_n((y_1,y_2,\cdots,y_n))) = (y_1,y_2,\cdots,y_n).$$

3. $$f$$ is injective.

This was the most challenging part for me. Suppose that $$f$$ is not injective so that there exists $$x,y \in H$$ such that $$x\neq y$$ and $$f(x) = f(y).$$ Motivated by Daniel Fischer's comment(Link), I set $$G = \ker f$$ and $$g_i= {\tau_i}^{'}$$ for all $$i$$, where $${\tau_i}^{'} = \tau_i|_G$$. Note that each $${\tau_i}^{'}$$ is a homomorphism. After that, define $$u_1,u_2 : G \rightarrow H$$ as $$u_1(g) = g$$ and $$u_2(g) = e_H.$$ Clearly, these two functions are homomorphisms. Then, $$(\tau_i\circ u_1)(g) = \tau_i(g) = e_{G_i} = \tau_i(e_H) = \tau_i(u_2(g)) = (\tau_i\circ u_2)(g).$$ It follows that $$\tau_i\circ u_1 = \tau_i\circ u_2 = {\tau_i}^{'}$$ for every $$i.$$

Since $$f(x) = f(y), f(xy^{-1}) = f(x)f(y^{-1}) = f(x)f(y)^{-1} = (e_{G_1},e_{G_2}, \cdots,e_{G_n}).$$ Thus, $$xy^{-1} \in \ker f = G.$$ Also, as $$x \neq y,$$ $$xy^{-1} \neq e_H.$$ Then, $$u_1(xy^{-1}) = xy^{-1} \neq e_H = u_2(xy^{-1}).$$ Thus, $$u_1 \neq u_2$$, which is a contradiction to the uniqueness hypothesis!

By 1,2,3, $$f$$ is an isomorphism. Thus, $$H\cong G_1\times G_2\times \cdots \times G_n.$$

Is my argument correct? Especially , part 3 was pretty tricky to me.

## 1 Answer

I haven't had time to go through all the details of your proof.

But here are some thoughts on the problem. The property you described in the beginning is an example of what in category theory is termed a universal property. Universal properties define things up to a unique isomorphism.

The universal property you described is also enjoyed by the projection maps from the product group $$G_1×\dots×G_n$$. Then using the uniqueness property, we can conclude that $$H$$ is indeed isomorphic to that product.