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Say I have a set S w/ an associative binary operation *: S x S -> S and a two-sided identity e, and let sES. Let SsubLL and SsubRR be elements of S such that Ssib:L * s = e = s * SsubRR

How can I prove that Ssib:L = SsubRR ?

Since the left and right identity are equal to e which is a two-sided identity, that proves that the left and right are equal just by the definition of a two-sided identity right? Is there some more in-depth proof I'm not seeing here?

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    $\begingroup$ Does this hold for all $s \in S$, or do $\tilde{s}_L$ and $\tilde{s}_R$ depend on $s$? (Note: this was typed as Does this hold for all $s \in S$, or do $\tilde{s}_L$ and $\tilde{s}_R$ depend on $s$?) $\endgroup$ – pjs36 Sep 8 '15 at 22:35
  • $\begingroup$ I thought they were the same thing. I'll be honest, I don't know what the squiggly line over the s stands for. But I figured since s is an element in S then $\tilde{s}_L$ and $\tilde{s}_R$ are as well. $\endgroup$ – pfinferno Sep 8 '15 at 23:10
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$$\tilde s_R=e*\tilde s_R=(\tilde s_L*s)*\tilde s_R=\tilde s_L*(s*\tilde s_R)=\tilde s_L*e=\tilde s_L\;.$$

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  • $\begingroup$ I had something like this but I didn't understand it. Could you elaborate a little bit? Thank you. $\endgroup$ – pfinferno Sep 9 '15 at 2:52
  • $\begingroup$ @pfinferno: Would be a lot more efficient if you pointed out which step(s) or aspect(s) you don't understand. $\endgroup$ – joriki Sep 9 '15 at 7:41
  • $\begingroup$ Well what does the tilde above the s mean? The only place I saw it online was a .pdf that this problem came from and it didn't give a description. Once I know what that is I think it will make sense. $\endgroup$ – pfinferno Sep 9 '15 at 15:17
  • $\begingroup$ @pfinferno: I'd suggest that when you use notation in a question that you don't understand, you should point that out -- it would make it easier to understand where your problems lie, and what sort of answer might be helpful. I don't think this tilde is a standard notational convention. I suspect it's being used to mark inverses. In any case $\tilde s_L$ and $\tilde s_R$ are left and right inverses of $s$, respectively, and the statement being proved is that with a two-sided identity left and right inverses are equal. $\endgroup$ – joriki Sep 9 '15 at 15:21

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