Image of subgroup and Kernel of homomorphism form subgroups Is my proof ok?
Let $f:G\to G^{\prime}$ be a group homomorphism and let $H\leq G$. $\text{Im}(H) = \{f(x):x\in H\}$. To show that $\text{Im}(H)$ is a group, it suffices to show that $f(x)f(y)^{-1}\in \text{Im}(H)$.
$$f(x)f(y)^{-1}=f(xy^{-1})\in \text{Im}(H)\text{ because } xy^{-1}\in H$$
$\text{Ker}(f)=\{x:f(x)=e^{\prime}\}$ ($e^{\prime}$ is the identity in $G^{\prime}$).
$$f(xy^{-1})=f(x)f(y)^{-1}=e^{\prime}(e^{\prime})^{-1}=e^{\prime}\in \text{Ker}(f)$$.
 A: It's ok but don't forget to show that $\mathrm{Im}(H)\ne \emptyset$ since $f(e)=e'\in\mathrm{Im}(H)$ and the same thing for $\ker f$.
A: Yes, your proofs are correct. To be perfectly rigorous you also need to say that these sets are nonempty (which is obvious, but still).
A: Reference: http://www.proofwiki.org/wiki/Group_Homomorphism_Preserves_Subgroups
Duplicates: If $K$ is a subgroup of $G$, then $\phi(K) = \{ \phi(k) | k \in K \}$ is a subgroup of $\bar{G}.$ Vice-versa?
Why is the image of a homomorphism a subgroup of the codomain?
Showing the image of a subgroup is a subgroup.
== Theorem ==
Let $\left({G_1, \circ}\right)$ and $\left({G_2, *}\right)$ be [[Definition:Group|groups]].
Let $\phi: \left({G_1, \circ}\right) \to \left({G_2, *}\right)$ be a [[Definition:Group Homomorphism|group homomorphism]]. 
Then:
: $H \le G_1 \implies \phi \left({H}\right) \le G_2$
where $\le$ denotes [[Definition:Subgroup|subgroup]]. That is, [[Definition:Group Homomorphism|group homomorphism]] preserves [[Definition:Subgroup|subgroups]].
Proof: Let $H \le G_1$. First note that from [[Null Relation is Mapping iff Domain is Empty Set]]:
: $H \ne \varnothing \implies \phi \left({H}\right) \ne \varnothing$
and so $\phi \left({H}\right)$ is not [[Definition:Empty Set|empty]].
Next, let $x, y \in \phi \left({H}\right) \iff \exists h_1, h_2 \in H: x = \phi \left({h_1}\right), y = \phi \left({h_2}\right)$
We want to use the the [[One-Step Subgroup Test]]. Hence need $x*y^{-1} \in \phi(H) :$
$x*y^{-1} = \phi(h_1)*[\phi(h_2)]^{-1} = \phi(h_1)*\phi(h_2^{-1}) = \phi(h_1\circ h_2^{-1}).$  
Since $H$ is a [[Definition:Subgroup|subgroup]], $h_1\circ h_2^{-1} \in H$. Hence $\phi(h_1\circ h_2^{-1}) \in \phi \left({H}\right) \le G_2$.
