You have a little problem with $\beta$, you never defined it.
I think you meant to say in $P$ that $\forall \beta(\alpha<\beta\implies...)$ and then at $\delta<\gamma<\beta$ it is for arbitrary $\beta$.
If this is what you meant you proof is correct.
Here is a different way, which avoid the use of proper class(apart from $V$).
Let $f:\alpha\to V,g:\gamma\to V, f(x)=G(f\restriction x), g(y)=G(g\restriction y)$
Assume $\gamma\in\alpha(\gamma<\alpha)$.
Let $\delta<\gamma$ satisfy the induction requirements, for all ordinals lesser than $\delta$ we have $f,g$ equal.
From assumption we have $f\restriction \delta=g\restriction \delta$
Thus $G(f\restriction \delta)=G(g\restriction \delta)$
But this is nothing more than $f(\delta)=g(\delta)$.
This means that $\forall\delta\in\gamma(\delta<\gamma)(f(\delta)=g(\delta))$ hence $f\restriction \gamma=g$.
This is exactly the same as yours apart from the fact that my $f,g$ are fixed so it is easier to follow in my opinion.
We want to prove the following
Theorem: For each ordinal $\alpha$, the following equalities characterize the operation $\xi\longmapsto\alpha+\xi$:
$$\begin{cases}
\alpha+0=\alpha\\
\alpha+S(\beta)=S(\alpha+\beta)\\
\alpha+\gamma=\text{sup}\{\alpha+\delta\;|\;\delta<\gamma\}\quad\text{if }\gamma\text{ is a limit ordinal}
\end{cases}$$
But first, let's prove the following:
Lemma: For any ordinals $\alpha,\beta$ and $\gamma$:
(a) $\alpha+0=0+\alpha=\alpha$
(b) $\alpha+1=S(\alpha)$
(c) $\alpha+(\beta+\gamma)=(\alpha+\beta)+\gamma$ (associativity)
(d) $\alpha+S(\beta)=S(\alpha+\beta)$
Demonstration:
(a) Let $\langle A,<_R\rangle$ be a well ordered set of ordinal $\alpha$. The only well ordered set of ordinal $0$ is the empty set $\varnothing$, so the ordered sum has as base set $A\cup\varnothing=A$ and its well-order is simply the well-order $<_R$ of $A$, so its ordinal is $\alpha$ and $\alpha+0=\alpha$. Similarly, we can obtain $0+\alpha=\alpha$.
(b) Let $\langle A,<_R\rangle$ be a well ordered set of ordinal $\alpha$ and let $\langle\{b\},<_S\rangle$ be a well ordered set of ordinal $1$ such that $b\not\in A$. The ordinal $\alpha+1$ of the ordered sum $\langle A\cup\{b\},<_{R\oplus S}\rangle$ is $S(\alpha)$, because
$$\langle A\cup\{b\},<_{R\oplus S}\rangle\cong\langle S(\alpha),\in_{S(\alpha)}\rangle$$
Via the isomorphism $f:A\cup\{b\}\longrightarrow S(\alpha)$ defined by: for each $a\in A\cup\{b\}$
$$f(a)=\begin{cases}
g(a)\quad\text{if }a\in A\\
\alpha\qquad\,\text{if }a=b
\end{cases}$$
Where $g$ is the isomorphism between the well ordered set $\langle A,<_R\rangle$ and $\alpha$.
(c) Let $\langle A,<_R\rangle,\;\langle B,<_S\rangle$ and $\langle C,<_T\rangle$ be well ordered sets of ordinals $\alpha,\beta$ and $\gamma$ respectively, such that $A,B$ and $C$ are pairwise disjoint. The ordinal $\alpha+(\beta+\gamma)$ is the ordinal of the well ordered set $\langle A\cup(B\cup C),<_{R\oplus(S\oplus T)}\rangle$, and the ordinal $(\alpha+\beta)+\gamma$ is the ordinal of the well ordered set $\langle(A\cup B)\cup C,<_{(R\oplus S)\oplus T}\rangle$. This well ordered set are not only isomorphic but identical, because $A\cup(B\cup C)=(A\cup B)\cup C$ and $<_{R\oplus(S\oplus T)}=<_{(R\oplus S)\oplus T}$, so their respective ordinals are equal, i.e. $\alpha+(\beta+\gamma)=(\alpha+\beta)+\gamma$
(d) $\alpha+S(\beta)=\alpha+(\beta+1)=(\alpha+\beta)+1=S(\alpha+\beta)$
For the last part of the theorem, we just have to prove the following:
Lemma: For any ordinals $\alpha$ and $\gamma$, if $\gamma$ is a limit ordinal, then
$$\alpha+\gamma=\text{sup}\{\alpha+\delta\;|\;\delta<\gamma\}$$
Demonstration: We must prove the two corresponding inequalities. First, for each $\delta<\gamma$, we have that $\alpha+\delta<\alpha+\gamma$ (cf. Hrbacek & Jech Introduction to Set Theory, page 120, lemma 5.4 (a)), so
$$\text{sup}\{\alpha+\delta\;|\;\delta<\gamma\}\leq\alpha+\gamma$$
On the other hand, let $\eta<\alpha+\gamma$. We shall prove that there exists some $\delta<\gamma$ such that $\eta\leq\alpha+\delta$. We have that $\eta<\alpha$ or $\alpha\leq\eta$. In the first case, we simply take $\delta=0$. In the second case, we know there exists some $\beta$ such that $\eta=\alpha+\beta$ (cf. Hrbacek & Jech Introduction to Set Theory, page 121, lemma 5.5), and since $\eta<\alpha+\gamma$, it results that $\beta<\gamma$. So taking $\delta=\beta$, we have that $\eta\leq\alpha+\delta$, and
$$\alpha+\gamma\leq\text{sup}\{\alpha+\delta\;|\;\delta<\gamma\}$$
For the other part of the equivalence, that is, supposing that we define the sum of ordinals recursively, and we want to prove that the sum of ordinals is the ordinal of the ordered sum of two well ordered disjoint sets, just take a look again at the book of Hrbacek & Jech, page 120, theorem 5.3, where this fact is proved in full detail.
Best Answer
Define $\phi: G \to G$ as $\phi(g)=g^m$. Then because $G$ is abelian, $\phi$ is a homomorphism. Note that $\ker(\phi) = H$ and $k \to Hk$ defines an isomorphism from $K \to G/H$ so $G/H \cong K$.