Normal subgroups and cosets

Question

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

Answer 1

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 $G$

Answer 2

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

Answer 3

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 $N$.

Answer 4

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. $\square$