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!
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.
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.