My question reads:
Let K be a subgroup of a group G index 2. Prove that N is a normal subgroup as follows:
a) If a is not in N, prove that the coset Na consists of all the elements of G that are not in N
b) For each a in G, prove that a^(-1)Na ⊆ N
I do not understand how to approach this problem, for part a since a is not in N then is it not just straightforward that Na does not have elements in N? I am not too sure where to proof part here begins.
I would like help on getting this proof started so I can finish it.
We'll do it procedurally. Do ask questions if you are dissatisfied.
So, first in part a, given $a \notin N$, we want to show that $Na$ consists of all elements that are not in $N$.
To do this, we have to show two things:
1) Every element in $Na$ is not in $N$. To do this, we let $x \in Na$, then $x = na$ for some $n \in N$.Suppose, by way of contradiction, we assume $x \in N$. Then, right-multiplying both sides by $n^{-1}$, we get $a = xn^{-1}$. Now, $x \in N$, $n^{-1} \in N$ as $N$ is a subgroup, so $a \in N$, giving a contradiction. Hence, $x \notin N$, so $Na$ contains elements which at least,are not in $N$.
2) Every element not in $N$, is in $Na$. To do this, note that $N$ is of index $2$ in $G$. Hence, the number of cosets of $N$ is $2$. Can we describe these cosets? One coset is $N$ itself (or $Ne$, if you like). The other coset, is just described by taking $a \notin N$, and forming $Na$. Because $N$ is of index two, these are the only two cosets. Since every element of $G$ must belong to a coset (by definition of cosets being a partition by equivalence classes of $G$), we have that $Na$ contains exactly all elements that are not in $N$.
Now, we can go to part $b$:Consider $aNa^{-1}$. If $a \in N$, then this just reduces to $N$, so $a^{-1}Na \subset N$.
If $a \notin N$, then note that $Na$ consists of all elements not in $N$. Similarly, $aN$ consists of all elements not in $N$ (same argument for left cosets). By the description of these sets, $aN = Na$, or $aNa^{-1} = N$. Taking inverse of both sides, $a^{-1}Na = N$.
Hence, this completes the proof that $N$ is a normal subgroup of $G$.