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A group homomorphism satisfies $\phi (ab)=\phi (a)\phi (b)$ for every $a,b\in G$. Its kernel of a group homomorphism is $K=\{g:\phi (g)=e_H\}$. The homomorphism equation gives $\phi (e_G)=e_H$ by cancellation, and $\phi (g^{-1})=\phi (g{)}^{-1}$. Therefore $K$ contains the identity and is closed under multiplication and inverses, so it is a subgroup.

For $k\in K$ and $x\in G$,Thus conjugation carries $K$ into itself, and applying the inverse conjugation gives equality: $K$ is a normal subgroup. If $K=\{e,\xi \}$, conjugation cannot send $\xi$ to the identity because it is injective. ConsequentlyThis proves directly the normal subgroup of order two is central principle for kernels, including the kernel facts needed below.

Lagrange's theorem states that, for a finite group $G$ and a subgroup $H$,To prove it, define an equivalence relation by $x\sim y$ when $x^{-1}y\in H$. Its classes are the left cosets $xH$, which therefore partition $G$. Left multiplication by $x$ gives a bijection $H\to xH$, so every coset has $|H|$ elements. Counting this partition gives the formula; in particular $|H|$ divides $|G|$. Applying the theorem to the cyclic subgroup generated by an element shows that its order of a group element divides $|G|$.

Write the given order-two normal subgroup as $N=\{1,z\}$. Conjugation must preserve $N$ and cannot send its nonidentity element to the identity, so $gz{g}^{-1}=z$ for all $g\in G$. Thus $z$ is in the center of a group: a normal subgroup of order two is central.

Choose $g\notin N$. By Lagrange's theorem, its order of a group element is $2$, $p$ or $2p$. It cannot have order two: since $z$ is central, the four distinct elements $1,z,g,zg$ would form a subgroup of order four, whereas $4$ does not divide $2p$ for odd $p$. If $g$ has order $2p$, it already generates $G$. Otherwise $g$ has order $p$, commutes with $z$, and $(gz{)}^k=1$ implies $g^k=z^{-k}\in ⟨g⟩\cap N=\{1\}$. Hence both $p$ and $2$ divide $k$, so $gz$ has order $2p$. In either case