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The following corollary is quite confusing to me.

Let $N$ be a normal subgroup of a group, $G$ and let $\bar{G}$ denote the set of cosets of $N$ in $G$. Let $\pi: G \rightarrow \bar{G}$ be the canonical homomorphism. Let $a_{1},...a_{k}$ be elements of $G$ such that the product $a_{1}...a_{k}$ is in $N$. Then $\bar{a_{1}}...\bar{a_{k}} = 1$.

The proof is that the product $p=a_{1}...a_{k}$ is in $N$, so $\pi(p) = \bar{p} = \bar{1}$. Since $\pi$ is a homomorphism, so $\bar{a_{1}}...\bar{a_{k}} = \bar{p}$.

I'm wondering why it equals 1.

NB: a bar over something indicates an element from the set of cosets... I think (ref. page 66, Algebra by Artin).

Thanks in advance for your help.

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$\bar a = \bar b$ iff $aN = bN$ iff $b^{-1}a \in N$. Apply this with $\bar p = \bar 1$. –  Jack Schmidt Aug 31 '12 at 18:51
    
Not sure why this result needs the title of corollary actually. The identity element of the group $G/N$ is the coset $N = N1,$ where $1$ denotes the identity element of $G.$ –  Geoff Robinson Aug 31 '12 at 18:53
    
@JackSchmidt Is there someplace I would be able to see a proof for that? @ GeoffRobinson The title of corollary was not chosen by me (it was from a textbook) –  achacttn Aug 31 '12 at 19:03
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@achattn: My comment was not directed against you- I just meant that I did no see why the book felt the need to state the result as a corollary, when it looks like a tautological statement to me. –  Geoff Robinson Aug 31 '12 at 19:07
    
@GeoffRobinson Ah ok. –  achacttn Aug 31 '12 at 19:09
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2 Answers

up vote 3 down vote accepted

Recall that the cosets of $N$ form a partition of $G$: for any $a,b\in G$, either $aN=bN$, or $aN\cap bN=\varnothing$.

Now let $p\in N$. $N$ is a subgroup of $G$, so $p^{-1}\in N$, and therefore $1=pp^{-1}\in pN\cap N$. Thus, $pN$ and $N=1N$ are not disjoint, so they must be equal: $pN=1N$, or, in the bar notation, $\bar p=\bar 1$.

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When $aN = bN$, does it mean that it can be $a.n_{1}$ and $b.n_{2}$ where $n_{1}$ is not necessarily equal to $n_{2}$ (and both are in $N$). Thanks Edit again: Ah thanks, I understand now. I'm new to algebra so it is difficult to constantly be holding all the theorems and propositions etc at the same time. –  achacttn Aug 31 '12 at 19:05
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@achacttn: $aN=bN$ means that $\{an:n\in N\}=\{bn:n\in N\}$. If $c\in aN=bN$ and $a\ne b$, then yes, there will be distinct $n_1,n_2\in N$ such that $c=an_1=bn_2$. –  Brian M. Scott Aug 31 '12 at 19:15
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(Presumably redundant response to questions in comments; Brian's answer already contains this)

$\bar a$ is by definition $aN$, the coset of $N$ containing $a$.

If $\bar a = \bar b$, then by definition $aN = bN$.

Guido's lemma is that $aN = bN$ if and only if $b^{-1}aN = N$ if and only if $b^{-1}a \in N$.

While simple, this is extremely useful.

Proof: If $aN = bN$ then every element of $aN$ is contained in $bN$, so in particular, $a\cdot 1 \in aN$ is also in $bN$, so there is some $n \in N$ such that $a \cdot 1 = b \cdot n$. Dividing on the left by $b$ gives $b^{-1} a = n \in N$ and $b^{-1} a N = n N = N$. Conversely if $b^{-1}a \in N$, then there is some $n \in N$ such that $b^{-1} \cdot a = n$, and then $a \cdot 1 = b \cdot n$ so that $b$ is in both $aN$ and $bN$. Since cosets are either equal or disjoint, $aN = bN$. $\square$

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Thanks very much for that. –  achacttn Aug 31 '12 at 19:30
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