A map $\phi:G\to G’$ is called a group homomorphism if
\[\phi(ab)=\phi(a)\phi(b)\]
for any elements $a, b\in G$.

Proof.

Let $e, e’$ be the identity elements of $G, G’$, respectively.
First we claim that
\[\phi(e)=e’.\]
In fact, we have
\begin{align*}
\phi(e)&=\phi(ee)=\phi(e)\phi(e) \tag{*}
\end{align*}
since $\phi$ is a group homomorphism.
Thus, multiplying by $\phi(e)^{-1}$ on the left, we obtain
\begin{align*}
&e’=\phi(e)^{-1}\phi(e)\\
&=\phi(e)^{-1}\phi(e)\phi(e) && \text{by (*)}\\
&=e’\phi(e)=\phi(e).
\end{align*}
Hence the claim is proved.

Then we have
\begin{align*}
&e’=\phi(e) && \text{by claim}\\
&=\phi(gg^{-1})\\
&=\phi(g)\phi(g^{-1}) && \text{since $\phi$ is a group homomorphism}.
\end{align*}

It follows that we have
\begin{align*}
\phi(g)^{-1}&=\phi(g)^{-1}e’\\
&=\phi(g)^{-1}\phi(g)\phi(g^{-1})\\
&=e’\phi(g^{-1})\\
&=\phi(g^{-1}).
\end{align*}
This completes the proof.

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[…] Suppose that we have [phi(a)=phi(b)] for some elements $a, bin G$. Then the properties of the homomorphism $phi$ imply that begin{align*} phi(a)phi(b)^{-1}=e’\ phi(a)phi(b^{-1})=e’\ phi(ab^{-1})=e’. end{align*} In the second step, we used the fact $phi(b)^{-1}=phi(b^{-1})$, which is proved in the post “Group Homomorphism Sends the Inverse Element to the Inverse Element“. […]

[…] In the second step, we used the fact $f(g_2^{-1})=f(g_2)^{-1}$, which is proved in the post “Group Homomorphism Sends the Inverse Element to the Inverse Element“. […]

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[…] Suppose that we have [phi(a)=phi(b)] for some elements $a, bin G$. Then the properties of the homomorphism $phi$ imply that begin{align*} phi(a)phi(b)^{-1}=e’\ phi(a)phi(b^{-1})=e’\ phi(ab^{-1})=e’. end{align*} In the second step, we used the fact $phi(b)^{-1}=phi(b^{-1})$, which is proved in the post “Group Homomorphism Sends the Inverse Element to the Inverse Element“. […]

[…] In the second step, we used the fact $f(g_2^{-1})=f(g_2)^{-1}$, which is proved in the post “Group Homomorphism Sends the Inverse Element to the Inverse Element“. […]