A group action of a group $G$ on a set $X$ is a map $\phi:\, G\times X \rightarrow X$ such that
$$
\phi(e,x) = x, \quad\text{and}\quad
\phi(g, \phi(h,x) ) = \phi(gh,x),
$$
for all $x\in X$ and $g,h\in G$ ($e\in G$ is the identity element). Usually, one simply writes $g\cdot x \equiv \phi(g,x)$.
Given such a group action, the stabilizer subgroup of an element $x \in X$ is defined as
$$
G_x := \left\{ g\in G \; | \; g\cdot x = x \right\},
$$
i.e. the subgroup which fixes $x$. Likewise, for any subset $Y\subset X$, we can define the stabilizer $G_Y$ as the intersection of all $G_x$ for $x\in Y$. That is, $G_Y$ is the subgroup of $Y$ which fixes $S$ point-wise.
Note that you can also reverse the construction and ask: Given a subgroup $H$ of $G$, what is the subset $X_H\subset X$ which is stabilized by $H$? This is exactly the subset of fixed points of $H$.
Here, we consider $X=\mathbb C^{2^n}$ and the unitary group $G=\mathrm U(2^n)$ with its natural action/representation by matrix multiplication. Then, the stabilized set of a subgroup $H<\mathrm U(2^n)$ is always a subspace, namely the common 1-eigenspace of all $h\in H$. For this to be non-empty, it is sufficient that all $h\in H$ commute (and that all $h\in H$ have an eigenvalue 1). Thus, we'll consider Abelian subgroups in the following.
Note that given a subspace $V\subset \mathbb C^{2^n}$, its stabilizer $\mathrm U(2^n)_V \simeq \mathrm U(V^\perp)$ is generally huge. Albeit, you always get away with a much smaller group $\mathcal{S}_V \subset \mathrm U(2^n)_V$ of order $|\mathcal{S}_V| = \dim V$ which nevertheless stabilizes $V$.
In the context of quantum codes, we call $V\subset \mathbb C^{2^n}$ the code space or simply code and want to regard it as an embedding of a $k$-qubit Hilbert space $\mathbb C^{2^k}$ into the $n$-qubit Hilbert space $\mathbb C^{2^n}$.
Hence, we have $\dim V = 2^k$, and we can always find a minimal stabilizer subgroup with $|\mathcal{S}_V|=2^k$.
For stabilizer codes, we restrict to $G=\mathcal{P}_n\subset \mathrm U(2^n)$, the $n$-qubit Pauli group, defined as
$$
\mathcal{P}_n = \left\{ i^k \sigma_1\otimes\dots\otimes\sigma_n \; | \; k\in\mathbb Z_4, \sigma_j \in \{\mathbb I, X, Y, Z \} \right\}.
$$
The stabilizer subgroups of stabilizer codes than have the form $\mathcal S = \langle P_1,\dots,P_k\rangle$, where $P_i\in \mathcal{P}_n$ are commuting and independent generators.
Finally, there's the additional condition that $-\mathbb I \notin \mathcal S$ (since the stabilized subspace would otherwise be $\{0\}$). This is equivalent to requiring that this subgroup contains only Hermitian operators.
To see this, note that any element of the Pauli group is either Hermitian or anti-Hermitian. Any Hermitian element $P$ has order two, $P^2 = PP^\dagger = \mathbb I$, as $P$ is also unitary. For an anti-Hermitian element, $(iP)^2 = - \mathbb I$, hence those have order four. This also shows that $\mathcal S$ cannot contain anti-Hermitian elements as otherwise $- \mathbb I \in \mathcal S$. This argumentation also shows that $|\mathcal S| = 2^k$.
