= Solution
The <order of a finite group> $G$ is its number of elements. <Lagrange's theorem> states that for a <subgroup> $K\leq G$,
$$
|G|=[G:K]|K|,
$$
so $|K|$ divides $|G|$. In particular the <order of a group element> divides $|G|$, because it is the order of its generated <cyclic subgroup>.
For $|G|=4$, if some element has order four it generates the whole group, which is therefore the <cyclic group> $C_4$. Otherwise every nonidentity element has order two. Choose two distinct ones $a,b$ and put $c=ab$. Cancellation shows $c\notin\{e,a,b\}$, so $c$ is the remaining element. Since $c^{-1}=c$,
$$
ab=c=c^{-1}=(ab)^{-1}=b^{-1}a^{-1}=ba.
$$
The other products follow: $ac=ca=b$, $bc=cb=a$, and $a^2=b^2=c^2=e$. Thus the second possibility is the <Klein four-group>. The two groups are not <isomorphic> because only $C_4$ has an element of order four. This proves the order-four case of the <classification of groups of order p squared>.
A <group homomorphism> $\Phi:G\to H$ obeys $\Phi(gh)=\Phi(g)\Phi(h)$. It maps identity to identity and inverses to inverses. Its <group kernel> $K=\{g:\Phi(g)=e_H\}$ contains $e_G$ and is closed under $ab^{-1}$, so the <subgroup test> makes it a subgroup. For $k\in K$ and $g\in G$,
$$
\Phi(gkg^{-1})=\Phi(g)e_H\Phi(g)^{-1}=e_H.
$$
Thus $gKg^{-1}\subseteq K$; applying the same argument to $g^{-1}$ gives equality. Hence $K$ is a <normal subgroup>.
Define $\overline\Phi:G/K\to H$ by $\overline\Phi(gK)=\Phi(g)$. If $gK=hK$, then $h^{-1}g\in K$, giving $\Phi(g)=\Phi(h)$, so this is well-defined. Coset multiplication shows it is a <group homomorphism>. Its kernel is just the identity coset, since $\Phi(g)=e_H$ means $g\in K$. It is therefore injective, and the assumed surjectivity of $\Phi$ makes it surjective as well. This proves the <first isomorphism theorem> here:
$$
\boxed{G/K\cong H.}
$$
In the <quaternion group>, the element $-1=i^2=j^2=k^2$ commutes with each of $i,j,k$, because an element commutes with its square. It then commutes with their negatives too, and with $\pm1$. The set $N=\{1,-1\}$ is a subgroup because $(-1)^2=1$, and it is central, hence a <normal subgroup>.
The four quotient <cosets> are $N,iN,jN,kN$. They square to $N$, and $(iN)(jN)=kN$. Reversed products have the same coset: for example $(ij)^{-1}=k^{-1}=-k$, while $j^{-1}i^{-1}=(-j)(-i)=ji$, so $ji=-k$. The same calculation gives $kj=-i$ and $ik=-j$. After signs are identified, these are precisely the products in the <Klein four-group>. Define
$$
\boxed{\Phi(\pm1)=e,\qquad\Phi(\pm i)=a,\qquad\Phi(\pm j)=b,\qquad\Phi(\pm k)=c.}
$$
This preserves all products, is surjective, and has kernel exactly $N$. The <first isomorphism theorem> therefore gives
$$
\boxed{Q_8/\{\pm1\}\cong V.}
$$
This is the <quaternion group modulo its centre>: the reverse-product relations also show that none of $\pm i,\pm j,\pm k$ is central, so $Z(Q_8)=N$.
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