Proving $H$ is a normal subgroup of $G$ Let $V$ be a non-empty set and $(SV,\circ)$ the symmetric group of $V$. Let $W \subseteq V$ a non empty subset of $V$. We define the next two subgroups of $SV$: $ G = \{ \sigma \in SV  \mid \sigma(W)=W  \} $ and $H = \{ \sigma \in SV\mid \forall w \in W: \sigma(w)=w \} $.
I have to prove that $H$ is a normal subgroup of $G$. But I have some trouble visualising which elements there are in $G$ and $H$. For me it looks like the condition $\sigma(W)=W$ is always true and $\sigma(w)=w$ is only true for the identical permutation. This looks too easy and trivial. So I'm asking is this true what I'm saying? 
Thanks in advance!
 A: Well, for example let us take the set $V = \{a,b,c\}$. Then there are six permutations in $(SV,\circ)$. A non trivial permutation is one sending $a \to b , b \to a$ and $c \to c$, for example. I am sure you can figure out the rest of the elements.
Now, take $W = \{a,b\}$. $G$, for this $W$ is defined as the permutations which keep $W$ in $W$.
For example, let $\sigma_1$ be defined as $a \to b, b\to a , c \to c$. We note that $\sigma_1(a) = b \in W$ and $\sigma_1(b) = a \in W$. Thus, $\sigma$ sends every element of $W$ to some other element in $W$. This need not be the same element : for example, $a,b$  are not going to themselves under $\sigma_1$, but we have $\sigma_1 \in G$.
An example of an element not in $G$ would be $\sigma_2$ sending $a \to b , b \to c, c \to a$,  because $\sigma_2(b) = c$ which is not in $W$, but $b \in W$, so $\sigma_2(W) \neq W$.  
However, if every element of $W$ is sent to itself, that is more special than every element going to another in $W$. For example, the identical permutation has this property, because it sends every element to itself,and  therefore those in $W$ as well. Such permutations come into $H$.
For an example where a non-identical permutation can be in $H$ : take the set $V' = \{a,b,c,d\}$ and $W = \{a,b\}$. Then the permutation $a \to a , b\to b , c \to d , d \to c$ is not identical, but it sends every element of $W$ to itself. Such a permutation would be in $H'$ (where $H'$ is the subgroup for this $V',W'$).
However, $a \to b, b\to a , c\to d , d \to c$ is not an element of $H'$, but rather of $G'$, because every element of $W'$ is not going to itself, but at least is going to another element of $W$.
The permutation $a \to b,b\to c , c \to a, d\to d$ is seen to belong to neither $H'$ nor $G'$.
From the above , I leave you to see that $H$ is a subset of $G$(we don't know if they are groups, so we don't use subgroup yet).

We will show that $G$ is a group, I leave $H$ to you.
It is enough to check that $G$ contains the identity, and is closed under inverse and multiplication.
Let $\sigma,\tau \in G$. We claim $\sigma \circ \tau \in G$. To see this, take $w \in W$. Then $\sigma \circ \tau(w) = \sigma(\tau(w))$. Now, $\tau(w) \in W$ by definition of $\tau \in G$, and then $\sigma(\tau(w)) \in W$ by definition of $\sigma \in G$ (andd because $\tau(w) \in W$). Thus, $\sigma \circ \tau \in G$, because every element of $W$ goes to another in $W$.
For inverse, we need to use $\sigma(W) = W$.  Let $w\in W$. Then, from the equality, $\sigma(W) = W $ so $w \in \sigma(W)$. By definition of $\sigma(W) = \{\sigma(v) : v \in W\}$, there is some $v \in W$ so that $\sigma(v) = w$. But then, $\sigma^{-1}(w) = v\ in W$ by definition of the inverse. Thus, $\sigma^{-1}$ keeps elements of $W$ in $W$, so inverse closure is done.
The identity is clearly in $G$. Thus $G$ is a group.
Work similarly for $H$, it is even easier.
A: $G \le \operatorname{Sym}(V)$ is the subgroup of the bijections on $V$ whose restriction to $W$ is onto $W$ (and then bijective on $W$), while $H \le G$ is the subgroup of the bijections on $V$ whose restriction to $W$ is $\iota_W$. Then, $\forall \sigma \in G, \forall \tau \in H$, we get:
$$(\sigma^{-1}\tau\sigma)_{|W}=\sigma^{-1}_{|W}\tau_{|W}\sigma_{|W}=\iota_W$$
so that $\sigma^{-1}\tau\sigma \in H$, and $H \unlhd G$.
A: Here is a slightly more conceptual way to view these groups:
Let $U$ be the complement of $W$ in $V$. Then there is a natural way to identify $G$ with $SW\times SU$, using the notation for symmetric groups from the OP.
With this identification in place, the subgroup $H$ can be seen as just the component $\{e\}\times SU$, which is normal since it is a direct factor.
