On P. 21 of Keith Conrad's notes on semidirect products, he made the following claim:
For a group $G$, let $H=Z(G)=Z$. ... Conversely, if $G\cong Z\times G/Z$ (any isomorphism at all), then $Z$ has a complementary subgroup in $G$.
I am curious on the construction of such complementary subgroup. Here is my attempt:
Let $f:Z\times(G/Z)\to G$ be an isomorphism, and let $K:=f(\{e\}\times(G/Z))\unlhd G$. I suspect $K$ is a complement of $Z$.
We know that such condition is true if and only if the restriction $\pi'$ of the quotient map $\pi:G\to G/Z$ to $K$ is an isomorphism. There is already one isomorphism from $G/Z$ to $K$: $$G/Z\xrightarrow{i}\{e\}\times(G/Z)\xrightarrow{f}K:\overline{a}\mapsto(e_G,\overline{a})\mapsto f(e_G,\overline{a}).$$ If we could prove that $\theta:=f\circ i$ is the inverse of $\pi'$, the proof would be complete. When such condition is true, we have for each $\overline{a}\in G/Z$, $$\overline{a}=\pi(\theta(\overline{a}))=\overline{f(e_G,\overline{a})}.$$ It seems that this condition is too strange to be true. Hope anyone could find a better or correct construction. Thanks.
This is not true. Note that it suffices to give an example of a group $H$ such that $H/Z(H)\cong H$ and $Z(H)$ is nontrivial, since then you can let $G=Z(H)^\mathbb{N}\times H$ and then $Z(G)=Z(H)^{\mathbb{N}}\times Z(H)\cong Z(H)^\mathbb{N}$ and $G/Z(G)\cong H$, but $Z(G)$ does not have a complement in $G$ since if it did then $G/Z(G)$ would be centerless.
An example of such a group $H$ is the orthogonal group $O(2)$. To prove this, note that $O(2)$ is a semidirect product $\mathbb{R}/\mathbb{Z}\rtimes \mathbb{Z}/2$ where $\mathbb{R}/\mathbb{Z}$ is the rotations and $\mathbb{Z}/2$ is generated by a reflection, which acts on $\mathbb{R}/\mathbb{Z}$ by $x\mapsto -x$. The center is $\{0,1/2\}\subset \mathbb{R}/\mathbb{Z}$ and the quotient $O(2)/Z(O(2))$ is then a semidirect product $\mathbb{R}/\frac{1}{2}\mathbb{Z}\rtimes\mathbb{Z}/2$. But $\mathbb{R}/\frac{1}{2}\mathbb{Z}\cong \mathbb{R}/\mathbb{Z}$ by multiplying by $2$ (and this isomorphism preserves the actions of $\mathbb{Z}/2$), so this gives an isomorphism $O(2)\cong O(2)/Z(O(2))$.
(It is true if $G$ is finite, since then the center of $Z(G)\times G/Z(G)$ can only have $|Z(G)|$ elements so it must be just the first coordinate.)