# proving in cosets

• Mar 15th 2011, 11:15 PM
deniselim17
proving in cosets
I have 2 questions on cosets.

Let $G$ be a group with subgroup $\langle h \rangle$.
Let $g \in G \backslash \langle h \rangle$.
If there exists $N \lhd G$ such that $[a,h] \notin N$, show that $g \notin \langle h \rangle N$.

I have tried this problem for a long time, but I still can't get a convinced proof.

Another question is if $aN=bN$, is it true that $a=b$?
• Mar 16th 2011, 02:52 AM
tonio
Quote:

Originally Posted by deniselim17
I have 2 questions on cosets.

Let $G$ be a group with subgroup $\langle h \rangle$.
Let $g \in G \backslash \langle h \rangle$.
If there exists $N \lhd G$ such that $[a,h] \notin N$, show that $g \notin \langle h \rangle N$.

I have tried this problem for a long time, but I still can't get a convinced proof.

I assume your "a" must in fact be "g", so $[g,h]\notin N\Longrightarrow g^{-1}h^{-1}gh\notin N$ , so if $g\in \langle h\rangle N$ then

$g=h^rn\,,\,r\in\mathbb{Z}\,,\,n\in N\Longrightarrow [g,h]=n^{-1}h^{-r}h^{-1}h^rnh=n^{-1}h^{-1}nh=[n,h]$ , but

this last rightmost expression is in N (why? Hint: normality), so we get a contradiction.

Another question is if $aN=bN$, is it true that $a=b$?

Not at all: $aN=bN\Longleftrightarrow b^{-1}a\in N$ and that's all.

For example, take $N=5\mathbb{Z}\,,\,a=2\,,\,b=7\Longrightarrow 2+N=7+N$

Tonio

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