Proving that a subgroup is normal When we want to prove that a subgroup is normal, we usually show that the left and right cosets are equal, right? But what if a group has so many elements, is not abelian, and the subgroup does not have index 2...how can we show that such a subgroup is normal? 
Thanks in advance 
 A: There are many ways of showing a subgroup is normal. One would be to show that the subgroup is the kernel of some homomorphism. There are some general theorems that guarantee a subgroup of a finite group to be normal, by some combinatorial argument. For instance, if $H$ has index $p$ and $p$ is the smallest prime that divides $|G|$, then $H$ must be normal. (This extends the easy result that a subgroup of index $2$ is normal). 
So, when given a particular subgroup and asked if it is normal you can try any of a rather limited number of tricks, or resort to (any of the equivalent) definitions of normality. 
A: One particular way is to choose a set of generators for the group and subgroup. One can do this for infinite groups, but I'll focus on the case of finite groups. 
A set of generators $A = \{a_1^{\pm}, a_2^{\pm}, \dots, a_n^{\pm}\}$ (where $\pm$ denotes the element and its inverse) for a group $G$ are elements $a_i \in G$ (and their inverses) such that when you take all possible ``words'' in $A$, you can get all the elements of $G$. In other words, any $g \in G$ can be written as $g = a_{i_1}a_{i_2}\dots a_{i_m}$, where $a_{i_k} \in A$ and you can have as many repetitions as you want, and each of the $a_{i_k}$ can refer to any $a_i \in A$ and its inverse.  
Let's say then that $A$ generates $G$ and that a subset $B$ generates a subgroup $H$, and you want to check whether that subgroup is normal. Let's also assume that the normalizer of $B$ in $G$ is $G$, i.e., $N_G(B) = \{g \in G : gB = Bg\} = G$ (* -- check note at bottom) -- this amounts to saying that $B$ is a ``normal subset''. Then, for any $g \in G$, $h \in H$, 
\begin{align*}
ghg^{-1} &= gb_1b_2\dots b_n g^{-1} \tag{because B generates H} \\
&= gb_1 g^{-1}g b_2 \dots g b_ng^{-1} \\
&= \beta_1\beta_2\dots \beta_n
\end{align*}
where the $\beta_i \in B$ since $B$ is ``normal''. Since $B \subset H$ you have shown that it suffices to normality of $H$ on a set of generators of $H$. You should show that it suffices to use elements $a_i \in G$ instead of all $g$. 
(*) check out the definitions of normalizer, centralizer and stabilizer. These will provide you with a way of thinking about groups that should guide the general ``philosophical'' answer to your question. 
