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I'm new to group theory, and trying to understand some basic concepts about cosets.

I found the following 5 lemmas and their proof was left "as an exercise for the reader" but I can't figure them out.

Let $H$ be a subgroup of $G$ and $gH$ be a left coset with $g$ as its representative. I want to show the following 5 lemmas are equivalent:

  1. $g_1H = g_2H$
  2. $Hg_1^{-1} = Hg_2^{-1}$
  3. $g_1H \subseteq g_2H$
  4. $g_2 \in g_1H$
  5. $g_1^{-1}g_2 \in H$

In particular, I want to show 1) implies 2) implies 3) implies 4) implies 5) implies 1) again.

I don't know how to get from 1) to 2). But it's obvious that 1) implies 3) but the other way around is nonobvious.

1) implies 4) since the identity is in $H$, so $g_2 \in g_2H \in g_1H$

4) implies 5) by left multiplication with $g_1$. Again, 5) to 1) baffles me since I can't get from the $\in$ to an $=$.

Can someone suggest approaches for showing these equivalences?

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  • $\begingroup$ For 1) implies 2) you just need to take inverses. This means you have 1) iff 2) so then you only need to show 1) implies 3) etc. If you have $g_1^{-1}g_2 \in H$ then that means $g_1^{-1}g_2H=eH$ where $e$ is the identity. $\endgroup$
    – Sam Weatherhog
    Nov 9, 2015 at 4:33
  • $\begingroup$ Oh, thanks. I guess it makes sense that $H^{-1} = H$ since $H$ is a subgroup so it's closed under inverses. $\endgroup$
    – user2193268
    Nov 9, 2015 at 4:35
  • $\begingroup$ Two comments. I wouldn't really call those "lemmas" as they're just statements; sometimes they're true, sometimes not, depending on the various $g_i$ and $H$. Second, that 3 implies 1 is obvious, if you remember that all cosets of $H$ have the same size. $\endgroup$
    – pjs36
    Nov 9, 2015 at 5:27
  • $\begingroup$ And as someone else here mentioned, by symmetry, we have both contains, so 3) can imply 1). What do you mean that sometimes they're not true? $\endgroup$
    – user2193268
    Nov 9, 2015 at 5:30
  • $\begingroup$ Nvm, I understand what you mean. If we assume 1) is true, then the rest are true. $\endgroup$
    – user2193268
    Nov 9, 2015 at 5:40

1 Answer 1

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3 $\Rightarrow$ 4 is immediate by symmetry of $3$, it is easier to show $g_1 \in g_2 H$, but since $e \in H$, then $g_1=g_1e \in g_1H \subseteq g_2H$.

You have 4 $\Rightarrow$ 5

5$\Rightarrow 1$ If $g_1^{-1}g_2\in H$, then $g_1^{-1}g_2=h$ for some $h\in H$. Then $g_2=g_1h$, so multiplying by $H$ we have $g_2H=g_1hH=g_1H$ since $hH=H$ for any $h\in H$.

I think you now have all of them $1 \Leftrightarrow 2 \Rightarrow 3 \Rightarrow 4 \Rightarrow 5 \Rightarrow 1$

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  • $\begingroup$ Thanks! I don't have a way directly from 2) to 3) but since 1) and 2) are equivalent, using 1) => 3) is good enough for me. Thanks for the proof of 5) => 1), that was what I was most stuck on. $\endgroup$
    – user2193268
    Nov 9, 2015 at 5:28
  • $\begingroup$ Let $g_1 h \in g_1H$. By 2, we know $hg_1^{-1}=h'g_2^{-1}$ for some $h' \in H$. Then by taking inverses of both sides, we have $g_1h=g_2h'^{-1}\in g_2 H$ as desired. $\endgroup$
    – CPM
    Nov 9, 2015 at 5:51

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