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First of all, I'm not a mathematican in any sense, I'm just curiuous sometimes. So please forgive me for not being rigorous in stating my actual problem.

Given a (non-abelian) group with its usual axioms, consisting of the set $G$ (lets call the group also $G$) and an operation $\cdot$, do I have to proof that from $$h\cdot g=e$$ follows that $$(h\cdot g)\cdot g^{-1}=e\cdot g^{-1}$$ , where $h,g \in G$ and $g^{-1}$ is the inverse element of $g$ and $e$ is the neutral element?

In principle this question is very similar to

Algebra: What allows us to do the same thing to both sides of an equation?

but if I got it right, a group doesn't need to be composed of "numbers". It can be anything, such as a rotation maybe or a far more abstract object I can't even imagine

So basically my question is, if the answer to the related question applies to any group in general or if I have to prove this somehow?

Thanks in advance and sorry for my poor english- and math-skills.

OD IUM
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  • Yes, it works also for "non numbers" (as per many similar question linked to the linked question); it is enough that the operation is defined between the objects you are dealing with, and it is always true that : "if $a=b$, then $(a \text { op } c = b \text { op } c)$. See rules for equality. – Mauro ALLEGRANZA Oct 21 '19 at 15:12
  • Yes! The inverse in a group is unique! Assume $x^{-1}=a^{-1}$. The definition of the inverse gives us $x^{-1} x = e$ and if $x^{-1}=a^{-1}$ that is equivalent to $xa^{-1} =e$ and multiplying with $a$ gives $x=a$. – cptflint Oct 21 '19 at 15:19
  • Doesn't even need to be a group. Applies to everything. If $a$ and $b$ are two names for the same thing, and "$manipulate()$" is a manipulation that has a consistent result determined by its input then $manipulate(a)=manipulate(b)$... Admittedly abstract mathematics can into great detail as to how we can define,interpret, and prove such simple concepts of say "a thing is itself" but that's fairly advanced and abstract. – fleablood Oct 21 '19 at 15:25
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    @cptflint although that is what the op was eventually trying to show, his/her question was much more basic. S/he was asking how we can multiply both sides by $a$ and if that is something that needs to be prove. If $M=N$ then do we know $Ma=Na$. – fleablood Oct 21 '19 at 15:28
  • @fleablood Mhhh, sure? If $g^{-1}$ isn't unique, that result won't be true. $g^{-1}$ ist just an element of the group which satisfies $g^{-1} g= e$. If you don't know, that this is a unique element, you can not interpret the multiplication of $g^{-1}$ with other elements than $g$. Or am I missing something here?

    I think the uniqueness is necessary for that.

     – cptflint Oct 21 '19 at 15:32
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    Subtle side issue: To say: if $a=b$ then $manipulate(a)=manipulate(b)$ requires that "$manipulate()$" is well-defined and consistent. I think some confusion comes from things such as. If $a=b$ then $a^2=b^2$ and $a0 =b^2$ is true. But if $a^2=b^2$ or $a0=b0$ then we can't* say $a=b$. But this is because $manipulate$ means "square it" or "multiply" by zero, is well defined and consistent. The concept of "find a number that when squared is equal to the input" or "find a number that when multiplied by zero is the input" are NOT well-defined and are inconsistent. – fleablood Oct 21 '19 at 15:35
  • That's what I am talking about. The multiplication by $g^{-1}$ is only well defined because of its uniqueness.

    Thx for your clarification!

     – cptflint Oct 21 '19 at 15:38
  • @cptflint. If $g^{-1}$ isn't unique then referring to something as "$g^{-1}$" doesn't make sense. If you have $M$ is something $N=M$ and $a$ is something you will always have $Ma = Na$. and if $M$ is $h\cdot g$ (whatever that is) and $h\cdot g = e$ (whatever that is) and $g^{-1}$ is something (whatever it is) you will have $(h\cdot g)\cdot g^{-1}=e\cdot g^{-1}$. – fleablood Oct 21 '19 at 15:40
  • In a group you have the "division" property: If $ac = bc$ then $a=b$. But i)you have to prove this by proving that "divide by $c$" is a consistent manipulation in a group (The proof is, by axiom/definition every $c$ has $c^{-1}$ and the operation $\cdot$ is a binary operation and the result follows) and ii) in other mathematical constructs this is not a consistent manipulation. For example in "clock" arithmetic where we return to $0$ every time we get to $12$. We have $42=8$ and $45=8$ (because in counting to $20$ we go back to $0$ at $12$) but $2\not=5$. "divide by 4" not defined! – fleablood Oct 21 '19 at 15:51
  • Okay, call it $i(g)$. When defining a group you need to force that there is at least one element $i(g)$ which satisfies $e=i(g)*g$. The uniqueness is just a consequence of that! And multiplying that with another element is only well defined when it is a unique element. And if you understand that, what you said is the next step and the answer to the given problem. But I agree: Uniqueness can be assumed when writing $g^{-1}$. That was my fault :) – cptflint Oct 21 '19 at 15:51

1 Answers1

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If we have, say, $$a=b$$ then for any function $f$, any function at all, we have $$f(a)=f(b)$$ For two values that are the same, inputting them into $f$ will give you the same output, because $f$, being a function, evaluates to a unique value on a given input. Multiplying by $g^{-1}$ on the right is a function. Calling this function $m_g$, we have $$m_g(x) = x\cdot g^{-1}$$ For your equation, we are saying that $$h\cdot g = e$$ so $$(h\cdot g)\cdot g^{-1} = m_g(h\cdot g) = m_g(e) = e\cdot g^{-1}$$ Therefore the manipulation is valid.

Matt Samuel
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