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Is it needed that if two sided identity exists for a binary operation (on a set), still need to check if inverse exists or not.

I request a reason for the same.

I came across this issue in exercise in Chap. $2$ in 'Problem C — Operations on a Two-Element Set'; in the book on Abstract Algebra by Charles Pinter.
The solutions there state that out of four operations (out of $16$), i.e. $O_1, O_6, O_7, O_9$, that have two-sided identity; only two operations , i.e. $O_6, O_9$, have inverse.

I am confused over the discrepancy of having a two-sided identity; yet no inverse. I mean why two-sided identity is not a sufficient condition for inverse to exist wrt a binary operation on a set.
All I know that a two sided identity is needed for a unique inverse for any element in a set wrt a binary operation.

Edit:
Seek formal reasoning; as say exists for why two sided identity is a must for inverse to exist. To, show it is not sufficient, but only a necessary condition for inverse to exist.
It means that need show that identity finding lacks certain conditions that are needed to find inverse. Or, in other words; identity finding operation is a subset of the Inverse finding operation.

Edit 2:
In my comment below to @JaapScherphuis, have concluded that the condition (additional) to have inverse is to satisfy $a^2=e$ for each non-identity element in the set. Please vindicate or contradict.

Edit 3:
My doubt is how is it possible for a binary operation on a set (with order $2$) to have two-sided identity, yet not have $x^2 =e$ for each non-identity element $x$ in the set.
The answer for me is in the tables constructed, as it shows the possibility.
But, would be more satisfied if got answer for such binops on sets with order $\ge 3$.

jiten
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    Addition on ${0,1,2,\ldots}$ has a two-sided identity, viz. $0$, but $1$ has no inverse. – Angina Seng Jul 24 '20 at 08:02
  • @AnginaSeng Cannot there be a reason possible at this beginner level? Or, should just take it as a fact. – jiten Jul 24 '20 at 08:04
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    Even at a beginner's level, one should not just assume that one desirable property (possessing a two-sided inverse) automatically entails another (existence of inverses) without proof. – Angina Seng Jul 24 '20 at 08:05
  • @AnginaSeng I hope then an answer is coming. – jiten Jul 24 '20 at 08:07
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    @jiten - What kind of proof do you need to be persuaded that there's no non-negative $n$ with $1+n=0$? – Malice Vidrine Jul 24 '20 at 08:12
  • @MaliceVidrine Nice, but seek formal reasoning; as say exists for why two sided identity is a must for inverse to exist. To, show it is not sufficient, but necessary condition for inverse to exist. Putting your words, it can be stated as: there may not exist for all elements in the set a corresponding inverse element. But, then it means that need show that identity finding lacks certain conditions that are needed to find inverse. Or, in other words; identity finding operation is a subset of the Inverse finding operation. – jiten Jul 24 '20 at 08:23
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    A counterexample should be proof enough that just having a two-sided identity is not sufficient. In for example $O_1$, it is trivial to check that $b$ is a two-sided identity because multiplying by $b$ does nothing: $(x,b)=x$, $(b,x)=x$ for both $x=a$ and $x=b$, and just as easy to see that multiplying on the right by $a$ cannot be invertible because $(x,a)=a$ regardless of what $x$ is, so you cannot recover what $x$ is from the output. The fact that $b$ is an identity does not restrict $a$ enough to be invertible. – Jaap Scherphuis Jul 24 '20 at 08:28
  • @JaapScherphuis Thanks a lot for that, your comment is very close (or, may be is) to stating reason for the requirement that finding inverse is more subtle (could not find a better word) operation than finding a two-sided identity. – jiten Jul 24 '20 at 08:35
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    Alternatively, if you have any non-invertible binary operation on a set $S$, you could extend it to a larger set $S\cup{e}$ by simply requiring $e$ to act as an identity, so $(e,e)=e$ and $(s,e)=(e,s)=s$ for all $s\in S$. Extending it in this way gives you an operation with a two-sided identity, but it does not magically make the operation invertible on the other elements in the set. – Jaap Scherphuis Jul 24 '20 at 08:43
  • @JaapScherphuis Please elaborate your last comment. Also, as for the earlier thread; I feel the answer lies in the fact (or, additional condition) that for any non-identity element a, there should exist $a^2=e$. – jiten Jul 24 '20 at 09:02

1 Answers1

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Let $e$ be the identity element.

\begin{array}{|c|c|c|} \hline & e & a \\ \hline e & e& a \\ \hline a & a & a \\ \hline \end{array}

is an example where there is no inverse though there is an identity.

Consider the set to have $3$ elements, notice that the definition of identity just determines the row and the column corresponding to $e$. It doesn't ensure that each row and each column must have $e$ appearing. I can fill in other entries in the lower right corner without using $e$.

\begin{array}{|c|c|c|c|} \hline & e & a & b \\ \hline e & e& a & b\\ \hline a & a & &\\ \hline b & b & &\\ \hline \end{array}

We are not given the information that it is a group or it has some further properties.

Also, $x^2=e$ is not a necessary condition. For example consider the additive group mod $3$.

Siong Thye Goh
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  • Request to know that how the answer at: https://math.stackexchange.com/a/3823202/424260states , that states $ab+cd=r^2(\operatorname{cis}(\alpha+\beta)+\operatorname{cis}(\gamma+\delta))$. As per the problem, $a = r (\operatorname{cis}(\alpha)), b = r (\operatorname{cis}(\beta)), c = r (\operatorname{cis}(\gamma)), d = r (\operatorname{cis}(\delta))$. I cannot understand how line $ab$ is given by product of modulus; which should be reserved only for product of two vectors (or complex points), & not for enoting a line (here, $ab$) that should be given by $r\operatorname{cis}(\beta -\alpha))$ . – jiten Sep 17 '20 at 15:30
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    I am confused, why do you describe $ab$ as a line? – Siong Thye Goh Sep 17 '20 at 16:14
  • Do you feel that the concept of vector for $ab$ will solve the problem faced by me? I feel not, $ab$ cannot be product of two vectors. It can be a difference of two vectors only; here as modulus being the same; it should be difference of arguments of $b-a$ multiplied by modulus. – jiten Sep 17 '20 at 16:36
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    I am very confused by your doubt, why are you associating it wiht a vector? while we can associate it as a vector? I do not understand your doubt. – Siong Thye Goh Sep 17 '20 at 16:37
  • Vectors $a, b$ are given by a common modulus $r$, and arguments given by $\operatorname{cis}(\alpha), \operatorname{cis}(\beta)$. So, the line joining the two points $A, B$ is given by difference of the two points (denoting vectors $a, b$). But, the state answer uses product of $a, b$ instead. – jiten Sep 17 '20 at 16:43
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    I don't even understand your question.... it's beyond my ability to clear your doubt until I undersrtand you. why are you talking about vectors when the question is asking about multiplication. Also don't use long comment threads.... especially on an irrelevant post. – Siong Thye Goh Sep 17 '20 at 16:59
  • Please help me with my post at: https://math.stackexchange.com/q/3839509/424260. It concerns permutations. Also, important topic for me. – jiten Sep 25 '20 at 02:48
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    not sure if I can understand the question, is the text available online? – Siong Thye Goh Sep 25 '20 at 06:35
  • Sorry for the delay. Also, nowhere is that page or book. The scanned images are at: https://i.stack.imgur.com/vtg24.jpg , https://i.stack.imgur.com/7iIxj.jpg – jiten Sep 25 '20 at 14:56
  • Also, want to add that the book is build up by questions only. The three questions 15, 16, 17 are as a group. – jiten Sep 25 '20 at 16:09