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**Statement:**

Let $\phi : G\rightarrow G'$ be a homomorphism of groups and let $g\in G$. Then:

(1) $\phi (e)$ is the identity of $G'$

(2) $\phi (g^{n}) = [\phi(g)]^{n}$ for all $n\in\mathbb{Z}$

(3) If $a\in G$, then $\phi (a^{-1}) = \phi (a)^{-1}$

(4) If $|g|$ is finite, then $|\phi (g)|$ divides $|g|$

(5) If $H\leq G$, then $\phi [H]\leq G'$

(6) If $K'\leq G'$, then $\phi ^{-1} [K'] \leq G$

(7) $Ker(\phi) \leq G$

**Proof:**

Let $\phi$ be a homomorphism of $G$ into $G'$. Then

Multiplying on the left by $\phi(a)^{-1}$, we see that $e'=\phi(e)$. Thus $\phi(e)$ must be the identity element $e'$ in $G'$. The equation

(2)shows that $\phi(a^{-1})=\phi(a)^{-1}$.

Turning to Statement(5),let $H$ be a subgroup of $G$, and let $\phi(a)$ and $\phi(b)$ be any two elements in $\phi[H]$. Then $\phi(a)\phi(b)=\phi(ab)$, so we see that $\phi(a)\phi(b)\in\phi[H]$;thus, $\phi[H]$ is closed under the operation of $G'$.The fact that $e'=\phi(e)$ and $\phi(a^{-1})=\phi(a)^{-1}$ completes the proof that $\phi[H]$ is a subgroup of $G'$.

Going the other way for Statement(6),let $K'$ be a subgroup of $G'$.Suppose $a$ and $b$ are in $\phi^{-1}[K']$. Then $\phi(a)\phi(b)\in K'$ since $K'$ is a subgroup. The equation $\phi(ab)=\phi(a)\phi(b)$ shows that $ab\in\phi^{-1}[K']$.Thus $\phi^{-1}[K']$ is closed under the binary operation in $G$.Also,$K'$ must contain the identity element $e'=\phi(e)$, so $e\in\phi^{-1}[K']$. If $a\in\phi^{-1}[K']$,then $\phi(a)\in K'$, so $\phi(a)^{-1}\in K'$. But $\phi(a^{-1})=\phi(a)^{-1}$, so we must have $a^{-1}\in\phi^{-1}[K']$.Hence $\phi^{-1}[K']$ is a subgroup of $G$.