Exact Short Sequences - Determine if Split Exact Let $F$ be a field. Let $G$ be the subgroup of $GL_3(F)$ consisting of the matrices
\begin{bmatrix} 1 & a & c\\ 0& 1 &b \\ 0& 0 & 1 \end{bmatrix}
and $Z$ is the center of $G$. Note: The center $Z$ is
\begin{bmatrix} 1 & 0 & z\\ 0& 1 &0 \\ 0& 0 & 1 \end{bmatrix}
and $K$ is the kernel of the surjection $G\rightarrow F$ with 
$$
\begin{bmatrix} 1 & a & c\\ 0& 1 &b \\ 0& 0 & 1 \end{bmatrix}\rightarrow a.
$$
Are the following two sequences below "Split Exact"? Give details.
a)  $\{ e  \}\rightarrow Z\rightarrow G\rightarrow \frac{G} 
{Z}\rightarrow \{ e \}$
b) $\{ e \}\rightarrow K\rightarrow G\rightarrow F\rightarrow  \{ e \}$
Any help would be appreciated. I know the maps out and in of each exact sequence is trivial. On a) I know $Z$ to $G$ is an inclusion, while I believe $G$ to $G/Z$ is a reduction mod $Z$. I am thinking the inclusion and reduction would make it an exact split, but not sure exactly what justifies this. I am having trouble on b) overall.
 A: For a), we shall prove a general result.  Let $Z$ be the center of an arbitrary group $G$.  Then, the short exact sequence $$\text{Id}\to Z\overset{\iota}{\longrightarrow} G\overset{\pi}{\longrightarrow} (G/Z)\to\text{Id}$$
splits (right-splits, to be precise) if and only if the factor group $G/Z$ has trivial center and $G$ is isomorphic to the direct product $Z\times (G/Z)$.  Here, $\text{Id}$ denotes the trivial group, $\iota:Z\to G$ is the canonical injection, and $\pi:G\to(G/Z)$ is the canonical projection.  The converse is easy to prove, so we shall verify the direct implication.
Suppose that the short exact sequence splits.  Let $\sigma:(G/Z)\to G$ be a section of $\pi:G\to(G/Z)$.  (Here, a section of a surjective group homomorphism $\phi:G_1\to G_2$ is a group homomorphism $s:G_2\to G_1$ such that $\phi\circ s:G_2\to G_2$ is the identity map $\text{id}_{G_2}$ on $G_2$.  Note that $s$ is necessarily an embedding, or an injective homomorphism.)    Write $H:=\text{im}(\sigma)$.  Then, $H$ is a subgroup of $G$ such that $H\cap Z=\text{Id}$ (because $\sigma$ is an embedding).  We also have $HZ=G$ because $\pi\circ\sigma$ is the identity map $\text{id}_{G/Z}$ on $G/Z$, so $H$ contains a complete set of representatives of $Z$-cosets in $G$.  Finally, $H$ is a normal subgroup of $G$.  This is because $Z$ is the center of $G$, so the elements of $Z$ commute with the elements of $H$.  Plus, $hH=H=Hh$ for all $h\in H$.  Since $Z$ clearly is normal in $G$, the Internal Direct Product Theorem tells us that $G$ is the internal direct product $Z\times H$.  Now, $H\cong (G/Z)$ must have trivial center as the elements in the center of $H$ belongs to $Z$, but we know that $H\cap Z=\text{Id}$.
Back to the original problem, we note that $(G/Z)\cong (F\times F)$, and the group $F\times F$ is abelian (whence it does not have trivial center).  Consequently, the short exact sequence $$\text{Id}\to Z\to G\to (G/Z)\to \text{Id}$$ does not split, and $G$ is not a semidirect product $Z\rtimes (G/Z)$.

For b), we shall find a section $\varsigma:F\to G$ of the surjective map $\varphi:G\to F$ defined in the OP's question. Set, for each $a\in F$,
$$\varsigma(a):=\begin{bmatrix}1&a&0\\0&1&0\\0&0&1\end{bmatrix}\,.$$
Clearly, $\varsigma$ is a well defined group homomorphism and the composition $F\overset{\varsigma}{\longrightarrow} G\overset{\varphi}{\longrightarrow} F$ is the identity map on $F$.  Thus, the short exact sequence $$\text{Id}\to K\overset{\subseteq}{\longrightarrow} G\overset{\varphi}{\longrightarrow} F\to\text{Id}$$ splits, and this proves that $G$ is a semidirect product $K\rtimes F\cong (F\times F)\rtimes_{\psi} F$, where $\psi:F\to\text{Aut}(F\times F)$ is given by
$$t\mapsto\big((y,z)\mapsto(y+tz,z)\big)$$
for all $t,y,z\in F$.
