# Group $G$ isomorphic to kernel of a homomorphism $\times$ a subgroup of the center

Let $$G$$ be a group, and $$C$$ a subgroup of $$G$$ which is contained in the center of $$G$$. Suppose there exists a group homomorphism $$f: G \longrightarrow C$$ such that $$f(c) = c$$ for all $$c \in C$$. Let $$K$$ denote the kernel of $$f$$. Prove that $$G$$ is isomorphic to $$K \times C$$.

I believe the route to go here is to use the following theorem: Let $$G$$ be a group with normal subgroups $$H,J$$ such that $$HJ = G$$ and $$H \cap J = \{e\}$$. Then $$G \cong H \times J$$.

As the kernel of a group homomorphism from $$G$$ to $$C$$, it's clear that $$K$$ is a normal subgroup of $$G$$. Furthermore, any subgroup of $$Z(G)$$ is normal in $$G$$, so $$C$$ is normal in $$G$$. It's left to show that $$KC = G$$ and $$K \cap C = \{e\}$$. How can I show these last two points ? It's not clear to me why the set $$\{kc | k \in K, c \in C\}$$ is all of $$G$$, or why there are no nonidentity elements that lie in both $$K$$ and $$C$$.

Any help would be appreciated.

Thanks!

• Noting that $f(g)=c_g\in C\ \&\ f(c)=c,\forall c\in C\Rightarrow( f\circ f)(g)=f(f(g))=f(c_g)=c_g=f(g),\forall g\in G,$ this task is a special case of Prove that $G=\ker(f)\times\operatorname{im}(f)$. Nov 20 at 8:22

No nonidentity element of $$C$$ is in the kernel because $$f(c) =c\neq e$$ for $$c\in C$$ with $$c\neq e$$. Thus $$K\cap C=\{e\}$$.
For $$g\in G$$, we have that $$f(g) =c\in C$$, so $$gc^{-1}$$ is in the kernel. Thus every element can be written as $$(gf(g)^{-1})f(g)$$, so $$G=KC$$.