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The theorem is stated as follows in the book:

Let $\phi:G\rightarrow G'$ be a group homomorphism, and let $H=Ker(\phi)$. Let $a\in G$. Then the set

$\phi^{-1}[\{\phi(a)\}] = \{x\in G | \phi(x)=\phi(a)\}$

is the left coset $aH$ of $H$, and is also the right coset $Ha$ of $H$. Consequently, the two partitions of $G$ into left cosets and into right cosets of $H$ are the same.

I'm trying to parse this statement and it's not clear to me what claim the author is trying to make at the very end when he says "Consequently, the two partitions of $G$ into left cosets and into right cosets of $H$ are the same." I'm under the impression that, in general, the left and right cosets are not always the same. Under what condition are they the same? Under the condition that you have a homomorphism?

Let me mention that at this point, we're not supposed to know what a normal subgroup is. The author introduces the idea of a normal subgroup 2 pages later.

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$\forall a\in G,aH=Ha$ is precisely the condition of being a normal subgroup: this shows kernels of homomorphisms are always normal subgroups. (In fact, the converse is true: every normal subgroup $H\trianglelefteq G$ is the kernel of some homomorphism, in particular the quotient map $G\to G/H$.) So you are right, left and right cosets are not generally the same. It should be clear what it means for two partitions of a set $S$ to be the same: each partition is a set of disjoint subsets of $S$ whose union is $S$, and two partitions are the same when they are the same set of subsets. –  anon Oct 2 '12 at 18:42
    
In response to your edit: you may not need to know what the definition of a normal subgroup is in order to do your homework or read your text, but normality is the answer to your question "under what condition are they the same?" (Although to nitpick, it is technically a potentially different situation for each left coset $aH$ to be the right coset $Ha$ versus each left coset $aH$ be some right coset $Hb$. I believe they have the same answer regardless, but the latter is more work.) –  anon Oct 2 '12 at 18:53
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If, because of normality, you know that $aH$ is also a right coset, then since $Ha$ is the only right coset that contains $a$, it follows as the night the day that $aH=Ha$. –  Lubin Oct 2 '12 at 19:48

3 Answers 3

The condition is that $H$ is the kernel of a group homomorphism, not just any random subgroup.

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If $N \subseteq G$ is a subgroup such that $gNg^{-1} = N$ for all $g \in G$, then $N$ is called a normal subgroup. See Wiki on that. This is equivalent to saying that any left and right cosets $gN = Ng$ for any $g \in G$ are the same.

So the statement actually just says that every kernel of a homomorphism is a normal subgroup.

This is easy to see since for all $h \in H = \mathrm{ker} (\phi)$, $a \in G$ you get: $$\phi (aha^{-1}) = \phi (a) \phi(h) \phi (a^{-1}) = \phi (a) \phi (a^{-1}) = 1,$$ so $aha^{-1} \in \mathrm{ker}(\phi) = H$. Therefore $aHa^{-1} \subseteq H$, multiplying by $a^{-1}$ from the left and $a$ from the right gives the other inclusion for its inverse. Since $a$ was arbitrary, equality follows.

Maybe you were confused by the definition of coset: I would read $aH$ as $\{ ah \in G;\; h \in H\}$, the set of all elements of the form $ah$ with $h \in H$. Similiary for $aHa^{-1}$. That both definitions given are equivalent, is done by checking: $$x \in aH \Leftrightarrow a^{-1}x \in H \Leftrightarrow \phi(a^{-1}x) = 1 \Leftrightarrow \phi(a) = \phi(x).$$

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What the statement is saying is that the kernel of a homomorphism is a normal subgroup.

I don't know if this clarifies or not, as I don't know the context: it could be that the author is trying to use this idea to introduce the notion of normal subgroup.

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