# Universal property of quotient group

Let $$H\le G$$ be a normal subgroup of $$G$$. Then we can think a natural projection map $$\pi:G \to G/H$$.

This map has the following universal property:

Let $$\phi:G \to G'$$ be a homomorphism. If $$H \subset \ker (\phi)$$, there is a unique homomorphism $$\tilde{\phi}:G/H \to G'$$ such that $$\phi=\tilde{\phi} \circ \pi$$.

I think it should be the universal property of the natural projection map. However, if the natural projection map $$q$$ is eligible to be called to have the universal property, then such map $$\pi$$ should be unique. That is,

Let $$f:G\to K$$ be a surjective homomorphism which has the following property: For any homomorphism $$\phi:G \to G'$$ satisfying $$H \subset \ker (\phi)$$, there exists a unique homomorphsim $$\tilde{\phi}:K\to G'$$ such $$\phi=\tilde{\phi} \circ f$$, then $$\ker (f)=H$$.

But I can't prove it. Would you let me know how to prove it? Or if it is not true, how can we modify the above universal property of natural projection map?

Any comment will be appreciated!

• Perhaps you want $\tilde{\phi}$ injective?
– lhf
Commented Jul 9, 2015 at 12:07
• @Ihf, Yes, that's exactly what I want to show! Commented Jul 9, 2015 at 12:30

Let $G = \mathbb Z,\ K = \mathbb Z/4$ and $H = 2\mathbb Z$. The projection map $f: \mathbb Z \to \mathbb Z/4$ has the property that for any map $\phi: \mathbb Z \to G'$ such that $2\mathbb Z\subset \ker(\phi)$ (which implies $4\mathbb Z \subset \ker(\phi)$), there is a unique map $\tilde\phi:\mathbb Z/4\to G'$ such that $\phi = \tilde\phi \circ f$. However, clearly the kernel of the map $\mathbb Z\to \mathbb Z/4$ is $4\mathbb Z$, not $H$.
The fix is to require that the map $f:G\to K$ has kernel containing $H$, i.e. $H\subset \ker(f)$. In that case, letting $G' = G/H$ in your statement, we get a unique map $K\to G/H$, and letting $G' = K$ in the universal property of quotient maps, we get a unique map $G/H\to K$. Composition of these two maps $K\to G/H$ and $G/H\to K$ must be identity because of the universal property in your statement. We conclude that $K$ and $G/H$ are isomorphic, and hence $\ker(G\to K) = H$, as desired.
• I am thank you for your kind reply. I didn't noticed that the category in which we are interested is the collection of morphisms $h:G\to G'$ such that $H \subset ker(h)$. If we think in this category, the natural projection is the initial object. Commented Jul 9, 2015 at 14:02