# Quotient group and normal subgroup

Assume N is a normal subgroup of G. Assume E is a subgroup of G/N. Thus E is a collection of right cosets of N in G. Let K be the set consisting of all the elements of G in the right cosets in E.

A)Prove that K is a subgroup of G that contains N. What is |K|?

B)Prove that K is normal in G if and only if E is normal in G/N.

I ran into this question and has a bit problem understanding "Let K be the set consisting of all the elements of G in the right cosets in E." What does it mean "cosets" there? Is it the cosets of K, which are in E?

• They are saying that $K$ is the union of the cosets of $N$ which make up $E$. Another way of looking at it, if $\phi:G\to G/N$ where $\phi:g\mapsto Ng$, then $K=\phi^{-1}(E)$. – Tim Raczkowski Mar 19 '15 at 3:54

Let's look at an example:

Suppose $G = \Bbb Z_{30}$ and $N = \langle 6\rangle = \{0,6,12,18,24\}$.

Then $G/N = \{N, N+1, N+2, N+3, N+4,N+5\}$

Let's take $E = \{N,N+2,N+4\} = \langle N+2\rangle$

Now $N+2 = \{2,8,14,20,26\}$ and $N+4 = \{4,10,16,22,28\}$, so:

$K = N \cup (N+2) \cup (N+4) = \{0,2,4,6,8,10,12,14,16,18,20,22,24,26,28\} = \langle 2\rangle$.

How does $E$ compare with $K/N$?

Now here, it is clear to see that $K$ is, in fact, a subgroup of $G$. But one example does not prove the general case.

So proving (A) involves 2 things: proving $K$ is a subgroup, and that $N \subseteq K$. Let's do the second part, first.

So pick any old element $n \in N$. Then $n$ is in the coset $N$, which is in EVERY subgroup of $G/N$, since $N$ is the identity of $G/N$. So $N \in E$, and thus $n$ is in some coset in $E$, so $n$ is in $K$. Thus $n \in N \implies n \in K$, so $N \subseteq K$.

Now let's suppose that $x,y \in K$. This means that $x \in Ng$ for some coset $Ng \in E$, and similarly $y \in Nh$, for some coset $Nh \in E$.

Since $E$ is a subgroup of $G/N$, it is closed under multiplication, thus $(Ng)(Nh) = Ngh \in E$.

Since $x \in Ng, x = ng$ and since $y \in Nh, y = n'h$ for some $n,n' \in N$. Since $N$ is normal, we have $Ng = gN$, so $ng = gn_1$ for some $n_1 \in N$. This means:

$xy = ngn'h = gn_1n'h$. Since $n_1n' \in N,\ gn_1n' \in gN = Ng$, so $gn_1n' = n_2g$ for some $n_2 \in N$.

Hence $xy = gn_1n'h = n_2gh \in Ngh$, and $Ngh \in E$, so $xy \in K$. This shows $K$ is closed under multiplication.

Finally, let $x \in K$. Then $x \in Ng$ for some $Ng \in E$. Since $E$ is a subgroup of $G/N$, we have $(Ng)^{-1} = Ng^{-1} \in E$. Now $x = ng$, so $x^{-1} = g^{-1}n^{-1} \in g^{-1}N$. Now $N$ is normal, hence $g^{-1}N = Ng^{-1}$, and we have shown if $x \in K$, that:

$x^{-1} \in Ng^{-1} \in E$, so $x^{-1} \in K$, and $K$ possesses all inverses. Thus $K$ is a subgroup of $G$ containing $N$.

If you know that $\phi:G \to G/N$ is a homomorphism, all this becomes MUCH easier.