We have already shown that A Direct Product Preserves Commutativity. Today, we will continue to work with commutativity as we show that commutativity is also preserved under homomorphisms.

**Homomorphisms**

Suppose that \(G\) is an Abelian group, \(H\) is a group and \(\alpha: G \to H\) is a group homomorphism. Then, \(\alpha(G)\) is a commutative subgroup of \(H\).

**Video Explanation**

In the video we show that isomorphisms preserve commutativity, but the proofs are extremely similar.

**Explaining Theorem**

In order to keep everything clear throughout the proof, we will state that \(\langle G, \cdot_{G} \rangle\) is a group with the operation \(\cdot_{G}\). Then \(\langle H, \cdot_{H}\rangle\) is a group with the operation \(\cdot_{H}\). Since \(G\) is commutative, we know that \(a \cdot_{G} b=b \cdot_{B} a\) for all \(a,b \in G\).

Then, we have to show that im\((\alpha)=\left\{c \in H: \alpha(a)=c \text{ for some } a \in G\right\}\) is a commutative subgroup of \(H\). In order to show that \(\alpha(G)\) is a subgroup, we can use the same theorem as in A Subgroup of Integers and show that \(c \cdot_{H} d^{-1} \in \alpha(G)\) for all \(c,d \in \alpha(G)\), where the inverse is as in \(H\).

Therefore, we would let \(c,d \in \alpha(G)\). This tells us that there is an \(a,b \in G\) such that \(\alpha(a)=c\) and \(\alpha(b)=d\). Then, since homomorphisms preserve inverses and multiplication, we find that

\begin{align*}

cd^{-1}&=(\alpha(a) \cdot_{H} \alpha(b)^{-1}) \\

&=(\alpha(a) \cdot_{H} \alpha(b^{-1})) \\

&=\alpha(a \cdot_{G} b^{-1}).

\end{align*}

Because \(G\) is a group, we know that \(a \cdot_{G} b^{-1} \in G\), so \(cd^{-1} \in \alpha(G)\) because \(\alpha\) maps the element \(a \cdot_{G} b^{-1}\) to \(cd^{-1}\).

Then, in order to show that it is a commutative group, we need to show that \(c \cdot_{H} d =d \cdot_{H} c\) for all \(c,d \in \alpha(G)\). We again have that \(c=\alpha (a)\) and \(d=\alpha(b)\) for some \(a,b \in G\). Here, we again use that homomorphisms preserve multiplication to get that

\begin{align*}

c\cdot_{H} d&= \alpha(a) \cdot_{H} \alpha(b) \\

&=\alpha( a \cdot_{G} b) \\

&=\alpha(b \cdot_{G} a) \\

&=\alpha(b) \cdot_{H} \alpha(a) \\

&=d \cdot_{H} c.

\end{align*}

Hence, \(\cdot_{H}\) is commutative within \(\alpha(G)\).

**Proof**

Let \(c,d \in \alpha(G)\). Then \(c=\alpha(a)\) and \(d=\alpha(b)\) for some \(a,b \in \alpha(G)\). Recalling that homomorphisms preserve multiplication and inverses, we then get that

\begin{align*}

c \cdot_{H} d^{-1}&=\alpha(a) \cdot_{H} (\alpha(b))^{-1} \\

&=\alpha(a) \cdot_{H} \alpha(b^{-1}) \\

&=\alpha(a \cdot_{G} b^{-1}).

\end{align*}

Since \(G\) is a group, \(a \cdot_{G} b^{-1} \in G\), hence \(\alpha(G)\) is a group.

Furthermore, note that

\begin{align*}

c \cdot_{H} d &=\alpha(a) \cdot_{H} \alpha(b) \\

&=\alpha(a \cdot_{G} b) \\

&=\alpha(b \cdot_{G} a) \\

&=\alpha(b) \cdot_{H} \alpha(a) \\

&=d \cdot_{H} c.

\end{align*}

Hence \(\alpha(G)\) is a commutative group. Therefore, homomorphisms preserve commutativity.

**Conclusion**

Now that we know homomorphisms preserve commutativity, we know that commutativity is preserved under the taking of homomorphisms, direct products and subgroups. That is, this equational identity is maintained by any of these three operations that we commonly use to create new groups from old. As we stated in A Direct Product Preserves Commutativity, this is not just limited to commutativity, but indeed will hold for any equational identity (again we will not show the general case). Knowing that these properties are invariant under the given operations helps us to understand a much larger family of groups.

As always, I hope this helps with your studying. If it did, make sure to share the post with other students that will find it helpful.