Check condition normal subgroup in these three examples Is the subgroup H of G is a normal subgroup of G, for:
$$ i)\ G = S_5, \ H = \{id, (1,2)\} $$
$$ ii) \ G = (Sym(\mathbb{N}), \circ), \ H = \{f\in Sym(\mathbb{N}) : f(0) = 0 \}$$
$$ iii) \ G = S_4, \ H = \{id, (1,2,3), (1,3,2) \} $$
I know, that subgroup is normal, if:
$$ \forall g\in G \ \ gH = Hg  \ \ - \ \ \ H \lhd G $$
But I don't know how. Could you solve example number one and two?
Thanks for your help. I know that $$ S_5 = \{id, (1,2,3,4,5), ..., (5,4,3,2,1)\} \ \ \ \ 5! = 120  $$ but I don't understand it.

While the condition for a subgroup being normal is correct, it is often not the most efficient or at least intuitive to check this via using it.
Multiplying by $g^{-1}$ you get equivalently $gHg^{-1} = H$ for all $g \in G$. 
So, to get you started you could just calculate $ghg^{-1}$ for a couple of $g\in G$ and $h \in H$ and see if all the elements are actually in $H$.
If you just find one where this is not true, you are done, and can conclude the group is not normal. 
If you do not seem to find such a counterexample you might start to suspect the subgroup is actually normal and then try to show that indeed for all $g \in G$ and $h \in H$ you have $ghg^{-1} \in H$. If you can show this you are also done and the group is a normal subgroup. Note: what I just said means $gHg^{-1} \subset H$ for all $g \in G$ (not equality). But this is yet another condition for a subgroup to be a normal subgroup.  
