# Normal subgroups and cosets

Let $N$ be a subgroup of a group $G$. Suppose that, for each $a$ in $G$, there exists $a, b$ in $G$ such that $Na=bN$. Prove that $N$ is a normal subgroup.

Attack: I found $b^{-1}N$ = $Na^{-1}$ but I am stuck!

Any help will be appreciated

• math.stackexchange.com/questions/283014/… May 4, 2013 at 19:04
• How are you defining a normal subgroup in this context? May 4, 2013 at 19:10
• Instead of "there exists $a, b$ in G" I think you might mean "there exists a $b$ in G" ... May 4, 2013 at 19:21

By hypothesis, every left coset of $N$ in $G$ is a right coset of $N$ in $G$. Thus, for $a\in G$, $aN$, being a left coset, must be a right coset. What right coset can it be?

Since $a=ae$ and obviously $e\in N$, you have that $a=ae\in aN$. So, whatever right coset $aN$ turns out to be, it must contain the element $a$. However, $a$ is in the right coset $Na$ and two distinct right cosets have no elements in common (remember that to be in a coset is an equivalence relation, which induces a partition of $G$). So this right coset is unique: $aN=Na$.

Hence, for every $g\in G$, you have $$gNg^{-1}=Ngg^{-1}=N$$ which is normality of $N$ in $G$

Hint: $bN$ contains $a$ and distinct left cosets of $N$ are disjoint.

One way of defining a Normal subgroup is that every right coset is a left coset - and that is what you are given.

Here you know that $a^{-1}Na=a^{-1}bN$

The left hand side contains 1. The right hand side is a coset of $N$ containing 1, so must be $N$.

Claim. For any $$b \in G, n_1 \in N$$, $$(bn_1)N = bN.$$ Proof. Let $$b \in G, n_1 \in N$$. Then \begin{aligned} (bn_1)N &= \{(bn_1)n_2 : n_2 \in N\} &&\quad\text{definition of left coset}\\ &= \{b(n_1n_2) : n_2 \in N\} &&\quad\text{associativity} \\ &= \{bn : n \in N\} &&\quad\text{since N is a subgroup} \\ &= bN, \end{aligned} as desired. $$\square$$

Main Proof. Let $$a \in G$$ be arbitrary. Note that $$e \in N$$, so $$a = ea \in Na$$. But $$Na = bN$$ for some $$b \in G$$, so $$a = bn_1$$ for some $$n_1 \in N$$. Since $$a = bn_1$$, $$aN$$ and $$(bn_1)N$$ are the same left cosets. By the above, $$aN = (bn_1)N = bN = Na.$$

Since $$Na = aN$$ for any $$a \in G$$, $$N$$ is normal. $$\square$$