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How do I go about doing this? I know that the definition of a quotient group is that it is the set of left cosets (aH) equipped with the following opeation:

$$ (aH)(bH) = (ab)H $$

Where the identity is $(1_G \cdot H) = H$.

How do I construct the group from here though? Or is this all the question is asking?

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  • $\begingroup$ I'd erase that second question about orbit, which isn't related at all to the first one, and would open a second thread for it. $\endgroup$ – DonAntonio Dec 7 '12 at 17:06
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You need to make (state) the connection between the hypothesis that $H\unlhd G$ and the conclusion that your $H/G$ is indeed a quotient group.

$$(aH)(bH) = (ab)H, \;\text {where the identity is}\;\; (1_G \cdot H) = H$$

You've got the pieces; you just need to connect them and verify that you do have a quotient group.

That is, simply verify that $(aH)(bH) = abH$ is well-defined, and that the operation defines a "group structure", to show that it is indeed a group.

This is easily established using the premise that $H\unlhd G$. For any old subgroup $H$ of $G$, it does not necessarily hold that $(aH)(bH) = abH,\;\; a, b \in G.\;\;$ It holds iff $H\unlhd G.$

Note: If you haven't learned that left coset multiplication on a subgroup $H$ is well defined if and only if $H$ is a normal subgroup of a group $G$, you should prove that. And if you haven't learned that the left (right) cosets of a normal subgroup $H$ of $G$ form a group $G/H$, then you need to prove that as well.

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  • $\begingroup$ Is that all I would say for the proof then? $\endgroup$ – Kaish Dec 7 '12 at 17:40
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    $\begingroup$ Yes, given that $H\unlhd G$, you know that $aHbH = abH$, which satisfies one requirement of a quotient group. Likewise, you know that $(1_G\cdot H = H)$. Hence, G/H is your desired quotient group! $\endgroup$ – amWhy Dec 7 '12 at 17:42
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That's is, you've constructed it! Now prove

(1) the operation is well defined, meaning

$$aH=a'H\,\,,\,\,\,bH=b'H\Longrightarrow (ab)H=(a'b')H$$

and

(2) the operation actually gives you a group structure on the set of (left or right, it doesn't matter) cosets of H.

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  • $\begingroup$ Is this important for the OP that he should care about the operation in abelian and non-abelian groups? I mean $aH$ and $a+H$. $\endgroup$ – mrs Dec 7 '12 at 17:08
  • $\begingroup$ Do I need to prove it if they're just asking for the construction? $\endgroup$ – Kaish Dec 7 '12 at 17:19
  • $\begingroup$ @Kaish, I think you must do that. $\endgroup$ – DonAntonio Dec 7 '12 at 17:31
  • $\begingroup$ @Babak, I don't think so. That'll come later. $\endgroup$ – DonAntonio Dec 7 '12 at 17:31
  • $\begingroup$ @Thanks, Don. Yes, it comes later. $\endgroup$ – mrs Dec 7 '12 at 17:45

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