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The book proves (a $\star$ b)$^{−1}$ = b$^{−1}$ $\star$ a$^{−1}$, where $\star$ is considered a binary group operation.

I will state the book's proof and then follow up with my questions.

Book's commentary:

In a group, to verify that an element h is the inverse of an element g, it suffices to show that g $\star$ h = e or h $\star$ g = e. In other words, we can prove that g $\star$ h = e $\rightarrow$ h $\star$ g = e and we can prove that h $\star$ g = e $\star$ g $\star$ h = e. For a proof that g$\star$h=e $\rightarrow$ h$\star$g=e, suppose that g$\star$h=e and k is the inverse of g. Then g$\star$k=k$\star$g=e.

Since g$\star$h=e and g$\star$ k=e, we have g$\star$h=g$\star$k. By multiplying by g$^{−1}$ on each side of this equation, and using associativity, the inverse property, and the identity property, we get h = k. So, h is in fact the inverse of g. Proving that h $\star$ g = e $\rightarrow$ g $\star$ h = e is similar.

If I do variable substitution, then for my problem I get the following proof:

Proof:

To verify that an element (b$^{−1}$ $\star$ a$^{−1}$) is the inverse of an element (a $\star$ b), it suffices to show that (a $\star$ b) $\star$ (b$^{−1}$ $\star$ a$^{−1}$) = e or (b$^{−1}$ $\star$ a$^{−1}$) $\star$ (a $\star$ b) = e. In other words, we can prove that (a $\star$ b) $\star$ (b$^{−1}$ $\star$ a$^{−1}$) = e $\rightarrow$ (b$^{−1}$ $\star$ a$^{−1}$) $\star$ (a $\star$ b) = e and we can prove that (b$^{−1}$ $\star$ a$^{−1}$) $\star$ (a $\star$ b) = e $\rightarrow$ (a $\star$ b) $\star$ (b$^{−1}$ $\star$ a$^{−1}$) = e.

For a proof that (a $\star$ b) $\star$ (b$^{−1}$ $\star$ a$^{−1}$)=e $\rightarrow$ (b$^{−1}$ $\star$ a$^{−1}$) $\star$ (a $\star$ b)=e, suppose that (a $\star$ b) $\star$ (b$^{−1}$ $\star$ a$^{−1}$)=e and k is the inverse of (a $\star$ b). Then (a $\star$ b) $\star$k = k$\star$ (a $\star$ b)=e.

Since (a $\star$ b) $\star$ (b$^{−1}$ $\star$ a$^{−1}$) = e and (a $\star$ b) $\star$ k = e, we have (a $\star$ b) $\star$ (b$^{−1}$ $\star$ a$^{−1}$)= (a $\star$ b)$\star$k. By multiplying by (a $\star$ b)$^{−1}$ on each side of this equation, and using associativity, the inverse property, and the identity property, we get (b$^{−1}$ $\star$ a$^{−1}$) = k. So, (b$^{−1}$ $\star$ a$^{−1}$) is in fact the inverse of (a $\star$ b). Proving that (b$^{−1}$ $\star$ a$^{−1}$) $\star$ (a $\star$ b) = e $\rightarrow$ (a $\star$ b) $\star$ (b$^{−1}$ $\star$ a$^{−1}$) = e is similar.

Question 1:

Are these the correct steps in the proof written out?

(a $\star$ b) $\star$ (b$^{−1}$ $\star$ a$^{−1}$)= (a $\star$ b)$\star$k

(a $\star$ b)$^{-1}$ $\star$ [(a $\star$ b) $\star$ (b$^{−1}$ $\star$ a$^{−1}$)]= (a $\star$ b)$^{-1}$ $\star$ [(a $\star$ b)$\star$k]=

[(a $\star$ b)$^{-1}$ $\star$ (a $\star$ b)] $\star$ (b$^{−1}$ $\star$ a$^{−1}$)= [(a $\star$ b)$^{-1}$ (a $\star$ b)]$\star$k=

e $\star$ (b$^{−1}$ $\star$ a$^{−1}$)= e $ \star $ k =

(b$^{−1}$ $ \star$ a$^{−1}$) = k

Question 2:

The structure of the proof looks like this to me:

(a $\star$ b) $\star$ [LHS inverse] = (a $\star$ b) $\star$ [RHS inverse]

[Another inverse]$\star$ (a $\star$ b) $\star$ [LHS inverse] = [Another inverse] $\star$ (a $\star$ b) $\star$ [RHS inverse]

e $\star$ [LHS inverse] = e $\star$ [RHS inverse]

[LHS inverse] = [RHS inverse].

If I already have 2 inverses (LHS inverse and RHS inverse) then what is the point in introducing the "Another inverse" in my boilerplate proof above? It seems redundant to me to have three separate inverses for cancellations. Is this the general proof pattern for these type of proofs?

Question 3:

The book states:

For a proof that (a $\star$ b) $\star$ (b$^{−1}$ $\star$ a$^{−1}$)=e $\rightarrow$ (b$^{−1}$ $\star$ a$^{−1}$) $\star$ (a $\star$ b)=e, suppose that (a $\star$ b) $\star$ (b$^{−1}$ $\star$ a$^{−1}$)=e and k is the inverse of (a $\star$ b). Then (a $\star$ b) $\star$k = k$\star$ (a $\star$ b)=e.

You want to prove: (a $\star$ b)$^{−1}$ = b$^{−1}$ $\star$ a$^{−1}$

How are you allowed to use what you are trying to prove as an assumption in the proof? If you want to show: (a $\star$ b)$^{−1}$ = b$^{−1}$ $\star$ a$^{−1}$, then how are you able to use this as a fact to prove the statement? This seems like supposing what you want to prove in your proof. This confuses me.

This is a screenshot from the text:

image

Xander Henderson
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geometric_construction
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  • You should state that $\star $ is a binary group operation. – fleablood Feb 10 '19 at 23:18
  • @fleablood - added. – geometric_construction Feb 10 '19 at 23:19
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    What book is that excerpted from? I'm surprised such an obfuscated proof would appear in any published book. – Bill Dubuque Feb 10 '19 at 23:24
  • This is an incredibly poorly written passage. I understand your confusion. – Servaes Feb 10 '19 at 23:24
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    Which book is this? Has the book proven that the identity element is unique? That inverses are unique? And that solely LH inverses or solely RH invers don't exist for groups? This proof seems to go into to much detail if those have been proven. But if not... well, we need to know what order things are proven. – fleablood Feb 10 '19 at 23:25
  • @fleablood none of what you asked has been proven. The book is: "Pure Mathematics for Beginners: A Rigorous Introduction to Logic, Set Theory, Abstract Algebra, Number Theory, Real Analysis, Topology, Complex Analysis, and Linear Algebra" by Steve Warner. The statement was more generalized than that, but I did variable substitution of the proof I was working on into that statement. This comes from the solutions manual. – geometric_construction Feb 10 '19 at 23:28
  • Okay.... I'm agreeing with everyone the proof is terrible. The proof is simple that $(ab)(b^{-1}a^{-1}) = (a(bb^{-1}))a^{-1}$ by associativity and that is $(ae)a^{-1} = aa^{-1} = e$ by definition of inverse and identity and $(b^{-1}a^{-1})(ab)=e$ is proven the same way. So $(b^{-1}a^{-1}) $ is the multiplicative inverse of $(ab)$ which you knew be definition must exist. Now there is another property to be proven that can assure you that inverses are unique. – fleablood Feb 10 '19 at 23:30
  • I'm suspecting the book the book is trying to make a point about the nature of proofs. To actually prove the result, these seems very inefficient. – fleablood Feb 10 '19 at 23:32
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    "A Rigorous Introduction to Logic, Set Theory, Abstract Algebra, Number Theory, Real Analysis, Topology, Complex Analysis, and Linear Algebra" ooh, I hate these books that are so narrow in scope.... :) – fleablood Feb 10 '19 at 23:33
  • The book proves it in a way like @Servaes showed, but it was kind of convoluted and I didn't fully understand the followup commentary. So, I took the commentary and did a direct variable substitution of my proof into it (which is what I posted), it didn't make much sense either, so hence my question. Thanks for clarifying this!! – geometric_construction Feb 10 '19 at 23:36
  • Please post the book's proof verbatim (or at least a screenshot). You should have noted that you transformed it. – Bill Dubuque Feb 10 '19 at 23:37
  • https://imgur.com/a/hHrgK7i – geometric_construction Feb 10 '19 at 23:39
  • The proof is trying to prove that since we know $(ab)^{-1} = k$ must exist that it does equal $(b^{-1}a^{-1})$ and nothing else. Its easy to prove that $(ab)(b^{-1}a^{-1})=e$ and then $(b^{-1}a^{-1})(ab)=e$ but ... maybe there are several other elements that do that? We don't know $(b^{-1}*a^{-1})=k$. So this proves that. – fleablood Feb 10 '19 at 23:43
  • Thanks for posting the original proof. You should unaccept the (non)answer so that you'll have better chance of attracting someone who can explain the issues with your translation. – Bill Dubuque Feb 10 '19 at 23:58
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    Hmm are you reading Bourbaki – Prince M Feb 11 '19 at 00:14

4 Answers4

2

This is not a direct answer to your questions, but hopefully helpful anyway. The passage states that

To verify that an element $(b^{−1}\star a^{−1})$ is the inverse of an element $(a\star b)$, it suffices to show that $$(a\star b)\star (b^{−1}\star a^{−1}) = e...$$

The rest of the passage is poorly written, confusing and irrelevant. Here's a clear proof:

By associativity $$(a\star b)\star (b^{−1}\star a^{−1})=a\star(b\star(b^{-1}\star a^{-1}))=a\star((b\star b^{-1})\star a^{-1}).$$ By definition of the inverse $b^{-1}$ of $b$ $$a\star((b\star b^{-1})\star a^{-1})=a\star(e\star a^{-1}).$$ By definition of the identity element $$a\star(e\star a^{-1})=a\star a^{-1}.$$ By definition of the inverse $a^{-1}$ of $a$ $$a\star a^{-1}=e.$$ This shows that indeed $$(a\star b)\star (b^{−1}\star a^{−1})=e.$$

Servaes
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Well, you need two show two things. First that $b^{-1}*a^{-1}$ is an inverse of $a*b$ and second that the inverse of an element is uniquely determined. Then it follows that $$(a*b)^{-1}=b^{-1}*a^{-1}.$$

In view of the first assertion: $$(a*b)*(b^{-1}*a^{-1}) = a*(b*b^{-1})*a^{-1} = a*e*a^{-1} = a*a^{-1}=e,$$ where $e$ denotes the unit element.

In view of the second assertion, let $b,c$ be inverses of $a$. Then $$b = b*e = b*(a*c) = (b*a)*c = e*c = c.$$ Done.

Wuestenfux
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Firstly, we need to show the the inverse of any element in a group $G$ is unique.

Suppose there are two inverses $$ b\ ,c, \ where\ b\neq c $$ such that $$ a*b=e=c*a.$$

Our first goal is to prove $$b=c.$$

So $$c\ =c*e = c*(a*b)=(c*a)*b=e*b=b.$$

Therefore the inverse of all elements in a group is unique.

Now we want to find the inverse of $$a*b.$$

Since $$a,b \in G,$$

it follows that $$ a^{-1},b^{-1} \in G. $$

Solving $(a*b)*(a*b)^{-1}=e$, or similarly $(a*b)^{-1}*(a*b)=e,$

just consider $$(a*b)*(a*b)^{-1}=e \\\implies a*(b*(a*b)^{-1})=e \ \ \ \ \ \ \ (by\ Assoc.\ law) \\\implies a^{-1}*(a*(b*(a*b)^{-1}))=a^{-1}*e \ \ \ \ \ \ \ (Multiply\ by\ a^{-1}) \\\implies (a^{-1}*a)*(b*(a*b)^{-1})=a^{-1}\ \ \ \ \ \ \ (by\ Assoc.\ law) \\\implies e*(b*(a*b)^{-1})=a^{-1} \\\implies b*(a*b)^{-1}=a^{-1} \\\implies b^{-1}*(b*(a*b)^{-1})=b^{-1}*a^{-1}\ \ \ \ \ \ \ (Multiply\ by\ b^{-1}) \\\implies (b^{-1}*b)*(a*b)^{-1}=b^{-1}*a^{-1}\ \ \ \ \ \ \ (by\ Assoc.\ law) \\\implies e*(a*b)^{-1}=b^{-1}*a^{-1} \\\implies (a*b)^{-1}=b^{-1}*a^{-1} $$

This is only one side of the proof; the other side is just similar to this.

J. W. Tanner
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Bello Bello
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Why not cut straight to the chase? Every group element has an inverse. Let $x$ be the inverse of $a \star b.$ Then $(a \star b) \star x = e.$ By associativity, $a \star (b \star x) = e.$ Compose both sides on the left with $a^{-1}$ to get $b \star x = a^{-1}.$ Repeat by hitting both sides on the left with $b^{-1}$ to get $x = b^{-1} \star a^{-1}.$ It's easy to verify (and in my opinion unnecessary) that $x$ is the inverse of $a \star b.$

Chris Leary
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