Find the order of a factor group $G/K$ Let $G$ be an abelian group and $g\in G$. Define a map $\phi:G\rightarrow G'$ when $\phi(x)=g^{-1}xg$. Let $K$ be the kernel of homomorphism $\phi$. Find the order of the factor group $G/K$.

I start to define $\phi$

Let $ϕ(x) = gxg^{−1}$ for all $x ∈ G$. Clearly $ϕ$ maps $G$ into $G$. First we will show $ϕ$ is a homomorphism. The homorphism condition holds for $ϕ$ because

$ϕ(xy) = g(xy)g^{−1} = gxyg^{−1} = gx(g^{−1}g)yg^{−1}=(gxg^{−1})(gyg^{−1})= ϕ(x)ϕ(y)$.

Next, $ϕ$ is injective, for if $ϕ(x) = ϕ(y)$ then $gxg^{−1} = gyg^{−1}$, so by multiplying this equation of the left of $g^{−1}$ and on the right by $g$ we get $g^{−1}(gxg^{−1})g = g^{−1}(gyg^{−1})g$, hence $(g^{−1}g)x(g^{−1}g) = (g^{−1}g) y(g^{−1}g)$, hence $x = y$. Finally, $ϕ$ is surjective because if $x ∈ G$ then $ϕ(y) = x$ where $y = g^{−1}xg$, since $ϕ(y) =ϕ(g^{−1}xg) = g(g^{−1}xg)g^{−1} = x$.

Am I fine in my definition?? Next:

Isomorphisms preserve order, so $|x| = |ϕ(x)| = |gxg^{−1}|$ for all $x ∈ G$. An
isomorphism is a bijection, and bijections preserve cardinality of sets, so for any subset $A ⊆ G, |A| = |gAg^{−1}|$.

Now how can I find the order of a factor group $G/K$??
I know
$$\frac{G}{ker\,\Phi}\cong \operatorname{Im}\,(\Phi )$$
let $x\in G$. It is clear $\Phi (gx{{g}^{-1}})=x$ then $G\subseteq G'$. On the other hand $\Phi$ is one to one so $G=G'$.

$\Phi$ is surjective, then
$$\frac{G}{K}\cong G$$
Edit
I know
$$\frac{G}{ker\,\Phi}\cong \operatorname{Im}\,(\Phi )$$
let $x\in G$. It is clear $\Phi (gx{{g}^{-1}})=x$ then $G\subseteq G'$. On the other hand $\Phi$ is one to one so $G=G'$.

$\Phi$ is surjective, then
$$\frac{G}{K}\cong G$$
Edit