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Problem 81 (The Image Of A Subgroup Under A Homomorphism Is A Subgroup)

Suppose that $f:G\to H$ is a homomorphism. Let $A$ be a subgroup of $G$. Show that $f(A)$ is a subgroup of $H$.

Solution

We will use the subgroup test to prove that $f(A)$ is a subgroup of $H$. Clearly $f(A) \subseteq H$. Since $A$ is a subgroup of $G$, we know $A$ is nonempty, and thus we can pick $x\in A$. Hence $f(x)\in f(A)$, and thus $f(A)$ is nonempty. Pick $y,z \in f(A)$. This means there exist $c,d\in A$ such that $f(c)=y$ and $f(d)=z$. We compute $yz=f(c)f(d)=f(cd)$. The last equality follows since $f$ is a homomorphism. Since $A$ is a group, we know $cd\in A$. It follows that $yz=f(cd)\in f(A)$.

Suppose $x\in f(A)$. We will show $x^{-1}\in f(A)$. We first show $e_H\in f(A)$. Since $A$ is a subgroup of $G$, we know $e_G\in A$. It follows that $f(e_G)=e_H\in f(A)$. Since $x\in f(A)$, there exists $z\in A$ such that $f(z)=x$. Since $z\in A$ and $A$ is a group, we know $z^{-1} \in A$. Hence $f(z^{-1})\in f(A)$. I claim $f(z^{-1})=x^{-1}$. We compute

$$ xf(z^{-1})=f(z)f(z^{-1})=f(zz^{-1})=f(e_G)=e_H $$

We also compute

$$ f(z^{-1})x=f(z^{-1})f(z)=f(z^{-1}z)=f(e_G)=e_H $$

By the subgroup test, we conclude that $f(A)$ is a subgroup of $H$.

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