# 5.3: Isomorphism Theorem

- Page ID
- 698

We've observed a few cases now where we: 1. Define a homomorphism \(\rho: G\rightarrow H\), and then 2. Notice that \(G/\mathord K \sim H\), where \(K\) is the kernel of \(\rho\). This isn't an accident!

Theorem 5.2.0: Isomorphism Theorem |
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Let \(\rho: G\rightarrow H\) be a homomorphism with kernel \(K\) and image \(I\). Then \(I\sim G/\mathord K\). |

The proof is just to build a correspondence between the cosets of the kernel \(gK\) and elements of the image \(I\). Indeed, in any coset \(gK\) all elements map to the same element of the image. \(\rho(gk)=\rho(g)\rho(k)=\rho(g)1=\rho(g)\) for any \(k\in K\).

This suggests a homomorphism from the set of cosets to the image: set \(\phi(gK)=\rho(g)\). This is a homomorphism, since \(\phi(ghK)=\rho(gh)=\rho(g)\rho(h)=\phi(gK)\phi(hK)\).

The map \(\phi\) is also one-to-one: if \(\phi(gK)=\phi(hK)\), we have \(\rho(g)=\rho(h)\), so that \(1=\rho(g^{-1}h)\), meaning \(g^{-1}h\in K\). Then \(h=g(g^{-1}h)\in gK\), which tells us that \(gK=hK\), since cosets are either equal or disjoint.

The map \(\phi\) is onto, since any element in the image may be written as \(\rho(g)\) for some \(g\), which is also the image of \(gK\) under \(\phi\). Therefore, the map \(\phi\) is an isomorphism.

**TODO: Pictures!**

This theorem is often called the "First Isomorphism Theorem." There are three isomorphism theorems, all of which are about relationships between quotient groups. The third isomorphism theorem has a particularly nice statement: \((G/\mathord N)/\mathord (H/\mathord N) \sim G/\mathord H\), which one can relate to the the numerical identity

\[\frac{ \frac{n}{m} }{ \frac{p}{m} }=\frac{n}{p}.\]

## Contributors and Attributions

- Tom Denton (Fields Institute/York University in Toronto)