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This is a question from the free Harvard online abstract algebra lectures. I'm posting my solutions here to get some feedback on them. For a fuller explanation, see this post.

This problem is from assignment $5$.

Prove directly that distinct cosets do not overlap.

Let $H$ be a subgroup of $G$ and let $a$, $b$, and $c$ be elements of $G$ such that $b\not\in aH$ and $c$ is in both $aH$ and $bH$. Then there are elements $h$ and $h^\prime$ in $H$ such that $c=ah$ and $c=bh^\prime$. So $ah=bh^\prime$ and $b=ahh^{\prime -1}$. But $hh^{\prime -1}\in H$ so $b\in aH$. This contradicts our original assumption. Therefore, there can be no element in more than one distinct coset.

Again, I welcome any critique of my reasoning and/or my style as well as alternative solutions to the problem.

Thanks.

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jobrien929
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  • That's the proof. I don't think you'll find an alternative solution, because it's immediate by appealing to the definitions. – Elchanan Solomon Jan 16 '12 at 00:33
  • Looks just fine to me. – Alex Becker Jan 16 '12 at 00:34
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    It looks fine. Another way of viewing this is to place an equivalence relation on $G$: $x \sim y$ if there exists an $h \in H$ such that $x = yh$. The equivalence classes are exactly the left cosets of $H$. You could also view this through group actions, but it doesn't seem like you're there yet. [I'm not claiming that these are essentially different.] – Dylan Moreland Jan 16 '12 at 00:37
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    Also, maybe it's good to be clearer as to why $b \notin aH$ implies that $aH$ and $bH$ are different. If you worry about that sort of thing then it seems cleaner to prove that $aH \cap bH \neq \varnothing$ implies $aH = bH$. – Dylan Moreland Jan 16 '12 at 00:43
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    I retract my (worthless) seal of approval from earlier. A lot of the right equations are there, but I'm concerned about this $b \notin aH$ thing. Perhaps the argument is: if the cosets are different then without loss of generality assume that $bH \not\subset aH$. So there exists an $h \in H$ such that $bh \notin aH$ and hence $b \notin aH$. – Dylan Moreland Jan 16 '12 at 01:37
  • On further review, I have to agree that introducing distinct cosets with $b\not\in aH$ is clumsy at best. But is it sufficient to simply say "assume $H$ is a subgroup of $G$ and $aH$ and $bH$ are distinct cosets "? – jobrien929 Jan 16 '12 at 02:09
  • The more I go back and read the statement of the problem, the more I wonder what the concept of "distinct cosets" means if you have to start out by assuming they might overlap.. This line of thought is leading me back to the way I originally worded the proof. – jobrien929 Jan 16 '12 at 02:25
  • @jobrien929: Since cosets are sets, "distinct" just means that one contains an element the other does not; that said, I think the cleanest way to do this is to proceed not by contradiction, but along the lines suggested by Dylan Moreland. – Arturo Magidin Jan 16 '12 at 06:02
  • @Arturo: Thanks for the shot of clarity. I was starting to think myself around in circles. So if I do want to start by saying $b\not\in aH$, then any set that contains $b$ is distinct from $aH$. Since $b\in bH$, $aH$ and $bH$ are necessarily distinct. Would that not be a good way to begin the proof? – jobrien929 Jan 16 '12 at 23:35
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    @jobrien929: I don't like it much; what if you encounter some coset "in the wild", as $bH$, but it's not $b$ that is not in $aH$? If you really want to start that way, it's much cleaner to say: "Let $aH$ and $bH$ be cosets, and assume that $aH\neq bH$; then either there exists $x\in aH-bH$, or $y\in bH-aH$; without loss of generality, assume that $x\in aH-bH$." And go from there. But, really, the cleanest way is to say: "Let $aH$ and $bH$ be cosets, and let $x\in aH\cap bH$". Then conclude from this assumption that $aH=bH$ as sets. – Arturo Magidin Jan 17 '12 at 00:08
  • @Arturo: Thanks again. I was definitely over-complicating things. – jobrien929 Jan 17 '12 at 00:43

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To prove the result at hand, you need to start with two distinct cosets sharing a common element, and prove that the cosets coincide. In saying that, "let $a$, $b$, and $c$ be elements of $G$ such that $b \not\in aH$ and $c$ is in both $aH$ and $bH$", you are assuming that $aH$ and $bH$ are distinct cosets with non-empty intersection and that doesn't mean that $b \not \in aH$ [as of now]. Also, $b \not \in aH$ does not also mean that, $bH \neq aH$ [as of now].

So, I think your proof is faulty here.

I'd fix it this way.

Suppose $c$ belongs to distinct cosets, $aH$ and to $bH$, say $$c = ah_1 = bh_2$$ where $h_1$, $h_2 \in H$. Then $a = bh_2h_1^{-1}$. Any element of $bH$ has the form $bh$ for some $h \in H$ and, $$ah = b(h_2h_1^{-1}h) \in bH$$

Since $h$ was arbitrary in $H$, we see $aH\subseteq bH$ .

The reverse inclusion, $bH \subseteq aH$, follows by a similar argument (using the equation $b = ah_1h_2^{-1}$ instead and arguing similarly).

This proves the claim.

There is a round about way of doing this, which I'll nevertheless mention here: The approach through equivalence relations.

Let $H \subseteq G$ be a subgroup of $G$. Define $\sim$ on $G$ by, $a\sim b$ iff there exists $h \in H$ such that, $a=bh$ for $a,b \in G$. Prove that your notion of cosets coincide with the equivalent classes of $\sim$. Now, you know that distinct equivalence classes are disjoint and hence your result.

Note that this leads you into Lagrange's Theorem in finite groups and in fact, the following is also true.

Let $H \subset G$ such that for all $a,b \in G$, $aH \cap bH=\emptyset$ or $aH=bH$ holds. Then $\exists g \in G$ such that $gH$ is a subgroup of $G$.

In case you have difficulty proving the above result, please let me know.

Hope this helps.

Edited to add @Dylan's viewpoint of this problem from group actions: Note that the equivalence relation we have defined can be viewed as a group action by looking at the definition of orbits. (i.e.) search for the action whose orbits are the equivalence classes of $\sim$. You'll identify that this action is the left multiplication action.

  • In the comments to the question itself I've started to wonder what "distinct cosets" means if you have to admit the possibility that they might overlap. When I first read Dylan's and your comments, I agreed with you but I think I've come full circle. The only way I can make sense of distinct is to have the element that "defines", for lack of a better word, the coset assumed to be outside the coset it's supposed to be distinct from. And I thought this was such a simple problem... – jobrien929 Jan 16 '12 at 02:39
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    @jobrien929 "Distinct" here should mean that you have two sets $aH$ and $bH$ and they are not equal. – Dylan Moreland Jan 17 '12 at 05:59