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.


  • $\begingroup$ That's the proof. I don't think you'll find an alternative solution, because it's immediate by appealing to the definitions. $\endgroup$ – Elchanan Solomon Jan 16 '12 at 0:33
  • $\begingroup$ Looks just fine to me. $\endgroup$ – Alex Becker Jan 16 '12 at 0:34
  • 3
    $\begingroup$ 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.] $\endgroup$ – Dylan Moreland Jan 16 '12 at 0:37
  • 3
    $\begingroup$ 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 $aH = bH$. $\endgroup$ – Dylan Moreland Jan 16 '12 at 0:43
  • $\begingroup$ 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 $b \notin aH$. $\endgroup$ – Dylan Moreland Jan 16 '12 at 1:37

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.

  • $\begingroup$ 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... $\endgroup$ – jobrien929 Jan 16 '12 at 2:39
  • 1
    $\begingroup$ @jobrien929 "Distinct" here should mean that you have two sets $aH$ and $bH$ and they are not equal. $\endgroup$ – Dylan Moreland Jan 17 '12 at 5:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.