I'm afraid the answer by Alexander is wrong. The given subgroup is indeed contained in the center of the semidirect product, but the reverse inclusion can fail.
Simple example: let $G_1$ be a center-free group, with an element $x$ of prime order $p$ (for instance, pick $G_1$ non-abelian of order 6 and $p=2$ or 3). Let the cyclic group $G_2$ of order $p$ act on $G_1$ so that the generator $c$ of order $p$ acts by conjugation by $x$: $\varphi(c^n)(g)=x^ngx^{-n}$. Then $\varphi:G_2\to\mathrm{Aut}(G_1)$ is injective, and hence the subgroup $(Z(G_1)\cap \mathrm{Fix}(\varphi))\times(Z(G_2)\cap\mathrm{Ker}(\varphi))$ is trivial. But the center of the semidirect product $G=G_2\rtimes G_1$ is not trivial: if the law of the latter is given by $(g_2,g_1)(h_2,h_1)=(g_2\varphi(g_1)(h_2),g_1h_1)$, then the element $(x,c^{-1})$ is central (thus the center $Z$ is cyclic of order $p$ and $G$ is direct product of $G_1$ and $Z$).
A correct description of the center of an arbitrary semidirect product should involve the kernel of the map $\varphi':G_2\to\mathrm{Out}(G_1)$, but it looks a bit complicated.
Edit: the center can be described as follows: let $f$ be the canonical map $G_2\to \mathrm{Inn}(G_2)=G_2/Z(G_2)$ and $s$ the canonical map $\mathrm{Ker}(\varphi')\to\mathrm{Inn}(G_2)$. Then the center of $G$ is the set of pairs $(g_2,g_1)\in G_2\rtimes G_1$ such that $g_1\in s^{-1}(f(\mathrm{Fix}(\varphi)))\cap Z(G_1)$, and $f(g_2)=s(g_1)^{-1}$.
Yes.
Remember what semidirect products do. When we take the direct product $G = Q \times K$, we get a group in which $Q$ and $K$ are both normal. Moreover, $G/Q = K$ and $G/K = Q$.
Semidirect products let us relax this to a case where only one of the groups ($K$, say) is normal, with $G/K = Q$, but $Q$ itself can sit inside $G$ however we want (If you're aware of the terminology, this is saying that we want to solve the extension problem in a way that the resulting sequence $1 \to K \to G \to Q \to 1$ is
split exact). I've chosen the letters $K$ and $Q$ so we can remember which group is the kernel and which is the quotient.
So we know we want to build a group $G$ containing $Q$ and $K$ with $G/K = Q$. Then since $K$ is normal, we know that $g^{-1}Kg = K$ for each $g \in G$. In particular for each $q \in Q$. It's easy to see that this is an automorphism of $K$, so we get a "conjugation action" $Q \to \text{Aut}(K)$ with $q \cdot k = q^{-1} k q$.
This means if we want to build a group $G$, we'll need to say what we want the conjugation action to be. What's slightly magical is that the conjugation action is the only data we need to pin down $G$!
Say $\varphi : Q \to \text{Aut}(K)$. Then we can find a (unique!) group $G = K \rtimes_\varphi Q$ so that
- $K$ and $Q$ are subgroups of $G$
- $G / K = Q$
- the conjugation action of $Q$ on $K$ agrees with $\varphi$. That is, $q^{-1}kq = \varphi(q)(k)$.
(If you prefer, we should really be writing $(1_K, q)$ and $(k, 1_Q)$ for the corresponding elements of $G$, but life is too short to not identify $K$ and $Q$ with their images in $G$)
So we see that, if $K = \langle k_i \mid R \rangle$ and $Q = \langle q_j \mid S \rangle$, we get a presentation
$$
G = K \rtimes_\varphi Q = \langle \ k_i, q_j \mid R, S,\ q_j^{-1} k_i q_j = \varphi(q_j)(k_i) \ \rangle
$$
which is exactly as you guessed!
I'll leave it as an (only slightly tedious) exercise that this presentation really does agree with the usual definition of semidirect product.
I hope this helps ^_^
Best Answer
You've got it wrong; $\mathbb Z_{55}^×$ has order $\varphi(55)=40$ and is isomorphic to $\mathbb Z_4×\mathbb Z_{10}$. The three non-direct semidirect products are derived from the three elements of order $2$ in $\mathbb Z_{55}^×$: $34$, $21$ and $-1$.