# 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.

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.