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Assuming your post should be prove $(a*b)^{-1}=b^{-1}*a^{-1}$ for $a,b \in G$ where $(G,*)$ is a group. Then by definition of inverses $(a*b)*(a*b)^{-1}=e$. Then using left inverse and associative properties we can get the following:
- User1520427
User1520427 - Group Theory - Proving $ (a*b)^ {-1} = (a^...
- Joseph Skelton
Joseph Skelton - Group Theory - Proving $ (a*b)^ {-1} = (a^...
- David Wheeler
David Wheeler - Group Theory - Proving $ (a*b)^ {-1} = (a^...
- Ragib Zaman
I am interested in mathematics and quantitative subjects in...
- User1520427
Prove that $a * b = a + b - ab$ defines a group operation on $\Bbb R \setminus \{1\}$
9 kwi 2015 · If $a>0$ and $b<0$, then $\gcd(a, -b)\cdot\operatorname{lcm}(a,-b)=|a\cdot(-b)|=|a\cdot b|$. Obviously, the set of the common divisors of $a$ and $b$ is equal to the set of the common divisors of $a$ and $-b$ .
22 cze 2017 · Let G be a finite group and let a and b be elements in the group. Then we prove that the order of ab is equal to the order of ba. Don't assume G is abelian.
Let G be a group and let a ∈ G. Prove that for any integers n,m ∈ Z, we have anam = an+m. [Suggestion: fix m ∈ Z and show that this works for n = 0 and n = 1, then prove it by induction on n for n ≥ 1.
6 lis 2016 · Recall that if the orders $m, n$ of elements $a, b$ of an abelian group are relatively prime, then the order of the product $ab$ is $mn$. (For a proof, see the post “Order of the Product of Two Elements in an Abelian Group“.)
First and foremost you have to prove that "$*$" is well defined, namely that: $$a,b\ne1\Longrightarrow a+b-ab\ne1$$ or, equivalently, that: $$a+b-ab=1\Longrightarrow (a=1)\vee(b=1)$$ And this is indeed the case, as: $$a+b-ab=1\iff a(1-b)=1-b$$ which splits into the two cases:
