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I've picked up on group theory again and I wanted to prove the cancellation rules for the elements in a group. But now I'm confused as to why my proof doesn't use circular reasoning:

Theorem (Cancellation laws). Let $G$ be a group. Then, $\forall a,g_1,g_2:$

  1. $ag_1=ag_2\implies g_1=g_2$
  2. $g_1a=g_2a\implies g_1=g_2$.

The proof I came up with and that I've checked online is correct, goes as following:

Proof of Theorem. Due to the group axioms, the implications $$ag_1=ag_2\implies a^{-1}(ag_1)=a^{-1}(ag_2)\implies(a^{-1}a)g_1=(a^{-1}a)g_2\implies1_Gg_1=1_Gg_2\implies g_1=g_2$$ and $$g_1a=g_2a\implies(g_1a)a^{-1}=(g_2a)a^{-1}\implies g_1(aa^{-1})=g_2(aa^{-1})\implies g_11_G=g_21_G\implies g_1=g_2$$ hold, therefore proving the theorem.

My question here is, why the first steph in the proof, nameley $ag_1=ag_2\implies a^{-1}(ag_1)=a^{-1}(ag_2)$, is not using circular reasoning, considering we want to prove that the cancellation laws are true? Does this step not rely on the condition that (left-)cancellation is holds true?

Thanks in advance for the answer(s)!

creativedisplayname
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    The first step only uses that $a^{-1}$ exists, so we can multiply it to both sides of an equation. – peek-a-boo Apr 04 '25 at 19:43
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    because you're using the converse implication, which holds for every operation in every algebra, by definition of operation. – amrsa Apr 04 '25 at 19:43
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    The point is that the cancellations property is only for finite groups equivalent to the existence of an inverse. One direction is always true, and trivial. If there exists $a^{-1}$, then of course you can cancel the $a$. – Dietrich Burde Apr 04 '25 at 19:54
  • Multiplication is well-defined, so if $x=y$ then $xz=yz$ and $zx=zy.$ – Thomas Andrews Apr 04 '25 at 20:06
  • By the group laws, $a^{-1}$ exists, and the group operation is a function from $G\times G\to G$. So, if $ag=ah$, then $a^{-1}(ag)=a^{-1}(ah)$ since functions send equations to equations. Then, associativity and the definition of inverse does the rest. – walkar Apr 04 '25 at 20:06
  • You've proved that ubiquitous fact that invertible elements are cancellable, by scaling by the inverse. Scaling by units (invertibles) is an invertible operation so always yields an equivalent equation - see the dupes. – Bill Dubuque Apr 04 '25 at 20:11
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    @peek-a-boo Indeed, we need only the existence of some inverse for $,a,$ in order to cancel $,a,$ as above (we don't need the uniqueness of such an inverse). This is often misunderstood, e.g. see here where this misunderstanding occurs in the popular algebra textbook by Dummit and Foote - when proving the uniqueness of solutions of $,ax=b,$ in a group. $\ \ $ – Bill Dubuque Apr 04 '25 at 20:43
  • The LHS and the RHS are equal, so doing the same things to them results in the same things. – Shaun Apr 04 '25 at 22:21

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