Here we will look at the composition of mappings and different properties preserved through these compositions. In particular, we will show that the composition of two isomorphisms results in an isomorphism. That is, the composition will preserve respect of operations, one-to-one and onto.

**Compositions of Isomorphisms**

Let \(G,H\) and \(K\) be groups. Furthermore, let \(\alpha:G \to H\) and \(\beta:H \to K\) be group isomorphisms. Then \(\beta \circ \alpha:G \to K\) is a group isomorphism.

**The Theorem**

With the theorem given, we want to make sure that we understand what is being stated. The first thing we want to recall is that an isomorphism is bijective homomorphism. That is, it is one-to-one and onto and respects the operation of the given group. That is, for \(\alpha\) we have the following,

- For all \(g \in G\), \(\alpha(g) \in H\).
- For all \(g_{1},g_{2} \in G\), we have that \(\alpha(g_{1} \cdot_{G} g_{2})=\alpha(g_{1}) \cdot_{H} \alpha(g_{2})\).
- For all \(h \in H\), there exists a \(g \in G\) such that \(\alpha(g)=h\).
- For any \(g_{1},g_{2} \in G\), if \(\alpha(g_{1})=\alpha(g_{2})\), then \(g_{1}=g_{2}\).

Furthermore, we would have the same properties for \(\beta\) if we replace the corresponding domains and codomain.

Then, we must show that \(\beta \circ \alpha: G \to K\) is an isomorphism. That is, we must show the same properties we had given above, using \(\beta \circ \alpha\) instead of \(\alpha\) and \(K\) instead of \(H\). Instead of trying to do all of these at once we will focus on one at a time.

**Well-defined**

Here we note that we proved that the composition of any two mappings, \(\gamma:A \to A\), \(\delta:A \to A\), is again a well-defined mapping. Despite having different domains and codomains, the fact that the composition is well-defined follows from an extremely similar proof. We will, therefore, give the adjusted proof below.

**Proof**

Let \(\alpha: G \to H\) and \(\beta:H \to K\) be mappings and \(g \in G\). Then note that, since \(g \in G\), we have that \(\alpha(g) \in H\) because \(\alpha\) is a well-defined mapping. Furthermore, since \(\alpha(G) \in H\) and \(\beta\) is a well-defined mapping, we get that \(\beta(\alpha(g)) \in K\). Hence, \(\beta \circ \alpha: G \to K\) is a well-defined mapping.

**Extra Note
**As an extra note, this proof goes further than needed for this problem. That is, it shows the more general case that the composition of any two mappings, \(\alpha\) and \(\beta\), if the codomain of \(\alpha\) is equal to the domain of \(\beta\), then the composition is well-defined from the domain of \(\alpha\) to the codomain of \(\beta\).

**Respects the Operation**

We now need to show the operation is respected for \(\beta \circ \alpha\). That is, we need to show that, if \(g_{1},g_{2} \in G\), then \(\beta( \alpha(g_{1} \cdot_{G} g_{2}))=\beta(\alpha(g_{1})) \cdot_{K} \beta(\alpha(g_{2}))\). Therefore, we will start with noting

\begin{align*}

\beta(\alpha(g_{1} \cdot_{G} g_{2}))&=\beta(\alpha(g_{1})\cdot_{H}\alpha(g_{2})),

\end{align*}

since \(\alpha\) is a homomorphism from \(G \to H\). Next, we have

\begin{align*}

\beta(\alpha(g_{1} \cdot_{G} g_{2}))&=\beta(\alpha(g_{1})\cdot_{H}\alpha(g_{2})) \\

&=\beta(\alpha(g_{1})) \cdot_{K} \beta(\alpha(g_{2})),

\end{align*}

since \(\beta\) is a homomorphism from \(H \to K\) and \(\alpha(g_{1}),\alpha(g_{2}) \in H\). We are now ready for our proof.

**Proof**

Let \(\alpha:G \to H\) and \(\beta:H \to K\) be group homomorphisms. Furthermore, let \(g_{1},g_{2} \in G\). Then

\begin{align*}

\beta(\alpha(g_{1} \cdot_{G} g_{2}))&=\beta(\alpha(g_{1})\cdot_{H}\alpha(g_{2})) \\

&=\beta(\alpha(g_{1})) \cdot_{K} \beta(\alpha(g_{2})).

\end{align*}

Hence \(\beta \circ \alpha:G \to K\) is a group homomorphism.

**Extra Note**

As we did last, we actually proved a more general case than we need and showed that the composition of two group homomorphism is again a group homomorphism.

**One-to-one**

Next, we need to show that the composition of these mappings is one-to-one. That is, if we suppose that

\begin{align*}

\beta (\alpha(g_{1})&=\beta(\alpha(g_{2})),

\end{align*}

for some \(g_{1},g_{2} \in G\), then \(g_{1}=g_{2}\). Here, we note that, because \(\beta\) is one-to-one and \(\alpha(g_{1}),\alpha(g_{2}) \in H\), we must have that

\begin{align*}

\alpha(g_{1})=\alpha(g_{2}).

\end{align*}

Then, since \(\alpha\) is one-to-one and \(g_{1}, g_{2}\) are in \(G\), we must have that \(g_{1}=g_{2}\) as required.

**Proof**

Let \(\alpha: G \to H\) and \(\beta:G \to H\) be one-to-one mappings and \(g_{1},g_{2} \in G\). Furthermore, if \)\beta(\alpha(g_{1}))=\beta(\alpha(g_{2}))\),

we have that \(\alpha(g_{1})=\alpha(g_{2})\) since \(\beta\) is one-to-one and \(\alpha(g_{1}),\alpha(g_{2}) \in H\). Now, \(g_{1}=g_{2}\), since \(\alpha\) is one-to-one and \(g_{1},g_{2} \in G\). Hence, \(\beta \circ \alpha\) is one-to-one.

**Onto**

As the last step in our proof, we need to show that \(\beta \circ \alpha\) is onto. That is, for every \(k \in K\), there exists a \(g \in G\) such that \(\beta \circ \alpha(g)=k\).

Here, we can note that since \(\beta\) is onto, there exists an \(h \in H\) such that \(\beta(h)=k\). Furthermore, since \(\alpha\) is onto, there exists a \(g \in G\) such that \(\alpha(g)=h\). Then, for this \(g\), we find that \(\beta(\alpha(g))=\beta(h)=k\).

**Proof**

Let \(\alpha:G \to H\) and \(\beta:H \to K\) be onto mappings of groups and \(k \in K\). Then, since \(\beta\) is onto, there exists an \(h \in H\) such that \(\beta(h)=k\). Furthermore, since \(\alpha\) is onto, there exists a \(g \in G\) such that \(\alpha(g)=h\). Hence, there exists a \(g \in G\) such that \(\beta(\alpha(g))=k\), so \(\beta \circ \alpha\) is onto.

**Final Proof**

We now have all the components we need to show that \(\beta \circ \alpha:G \to K\) is an isomorphism. Indeed, a formal proof would just involve placing the smaller proofs we have into one larger proof. Therefore, I will leave this to the reader.

**Conclusion**

Thank you for reading, and I hope this helped you while working with isomorphisms. If it did, be sure to like the post and share with someone else that may benefit from it.

As an extra note, this proof goes further than needed for this problem. That is, it shows the more general case that the composition of any two mappings, \(\alpha\) and \(\beta\), if the codomain of \(\alpha\) is equal to the domain of \(\beta\), then the composition is well-defined from the domain of \(\alpha\) to the codomain of \(\beta\).

**Respects the Operation**

We now need to show the operation is respected for \(\beta \circ \alpha\). That is, we need to show that, if \(g_{1},g_{2} \in G\), then \(\beta( \alpha(g_{1} \cdot_{G} g_{2}))=\beta(\alpha(g_{1})) \cdot_{K} \beta(\alpha(g_{2}))\). Therefore, we will start with noting

\begin{align*}

\beta(\alpha(g_{1} \cdot_{G} g_{2}))&=\beta(\alpha(g_{1})\cdot_{H}\alpha(g_{2})),

\end{align*}

since \(\alpha\) is a homomorphism from \(G \to H\). Next, we have

\begin{align*}

\beta(\alpha(g_{1} \cdot_{G} g_{2}))&=\beta(\alpha(g_{1})\cdot_{H}\alpha(g_{2})) \\

&=\beta(\alpha(g_{1})) \cdot_{K} \beta(\alpha(g_{2})),

\end{align*}

since \(\beta\) is a homomorphism from \(H \to K\) and \(\alpha(g_{1}),\alpha(g_{2}) \in H\). We are now ready for our proof.

**Proof**

Let \(\alpha:G \to H\) and \(\beta:H \to K\) be group homomorphisms. Furthermore, let \(g_{1},g_{2} \in G\). Then

\begin{align*}

\beta(\alpha(g_{1} \cdot_{G} g_{2}))&=\beta(\alpha(g_{1})\cdot_{H}\alpha(g_{2})) \\

&=\beta(\alpha(g_{1})) \cdot_{K} \beta(\alpha(g_{2})).

\end{align*}

Hence \(\beta \circ \alpha:G \to K\) is a group homomorphism.

**Extra Note**

As we did last, we actually proved a more general case than we need and showed that the composition of two group homomorphism is again a group homomorphism.

**One-to-one**

Next, we need to show that the composition of these mappings is one-to-one. That is, if we suppose that

\begin{align*}

\beta (\alpha(g_{1})&=\beta(\alpha(g_{2})),

\end{align*}

for some \(g_{1},g_{2} \in G\), then \(g_{1}=g_{2}\). Here, we note that, because \(\beta\) is one-to-one and \(\alpha(g_{1}),\alpha(g_{2}) \in H\), we must have that

\begin{align*}

\alpha(g_{1})=\alpha(g_{2}).

\end{align*}

Then, since \(\alpha\) is one-to-one and \(g_{1}, g_{2}\) are in \(G\), we must have that \(g_{1}=g_{2}\) as required.

**Proof**

Let \(\alpha: G \to H\) and \(\beta:G \to H\) be one-to-one mappings and \(g_{1},g_{2} \in G\). Furthermore, if \)\beta(\alpha(g_{1}))=\beta(\alpha(g_{2}))\),

we have that \(\alpha(g_{1})=\alpha(g_{2})\) since \(\beta\) is one-to-one and \(\alpha(g_{1}),\alpha(g_{2}) \in H\). Now, \(g_{1}=g_{2}\), since \(\alpha\) is one-to-one and \(g_{1},g_{2} \in G\). Hence, \(\beta \circ \alpha\) is one-to-one.

**Onto**

As the last step in our proof, we need to show that \(\beta \circ \alpha\) is onto. That is, for every \(k \in K\), there exists a \(g \in G\) such that \(\beta \circ \alpha(g)=k\).

Here, we can note that since \(\beta\) is onto, there exists an \(h \in H\) such that \(\beta(h)=k\). Furthermore, since \(\alpha\) is onto, there exists a \(g \in G\) such that \(\alpha(g)=h\). Then, for this \(g\), we find that \(\beta(\alpha(g))=\beta(h)=k\).

**Proof**

Let \(\alpha:G \to H\) and \(\beta:H \to K\) be onto mappings of groups and \(k \in K\). Then, since \(\beta\) is onto, there exists an \(h \in H\) such that \(\beta(h)=k\). Furthermore, since \(\alpha\) is onto, there exists a \(g \in G\) such that \(\alpha(g)=h\). Hence, there exists a \(g \in G\) such that \(\beta(\alpha(g))=k\), so \(\beta \circ \alpha\) is onto.

**Final Proof**

We now have all the components we need to show that \(\beta \circ \alpha:G \to K\) is an isomorphism. Indeed, a formal proof would just involve placing the smaller proofs we have into one larger proof. Therefore, I will leave this to the reader.

**Conclusion**

Thank you for reading, and I hope this helped you while working with isomorphisms. If it did, be sure to like the post and share with someone else that may benefit from it.