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I wonder if you would consider this proof of my hypothesis of

$(ab)^{-1} = a^{-1}b^{-1}$ with $a,b\ne0$ correct.

By definition of the inverse of $ab$: $$(ab)^{-1}(ab)=1$$

Multiplying the equation with $a^{-1}$ and $b^{-1}$ gives $$(ab)^{-1}aba^{-1}b^{-1} = 1a^{-1}b^{-1}=a^{-1}b^{-1}$$ using property $1x=x$. On the left hand side I rearrange the terms using the property $xy=yx$ to $$(ab)^{-1}aa^{-1}bb^{-1}= a^{-1}b^{-1}$$ which gives by using again the definition of the inverse $$(ab)^{-1}1\cdot1= a^{-1}b^{-1}$$ Using the property $1x=x$ on the left twice I have the answer.

Binomi
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    It depends on what $a$ and $b$ are here. If $a$ and $b$ are elements of a group, not necessarily abelian, then the correct generalization is $(ab)^{-1} = b^{-1} a^{-1}$. – Qiaochu Yuan Jan 13 '21 at 22:57
  • I second Qiaochu Yuan's comment. For emphasis: what are $a$ and $b$? – Shaun Jan 13 '21 at 23:29
  • This is sometimes called "socks and shoes". –  Jan 14 '21 at 00:31

3 Answers3

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Yes, this is correct. You could even do this without assuming commutativity, but then you would get $$(ab)^{-1} = b^{-1}a^{-1}.$$

Klaus
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You are correct, but this can be written more concisely as $(ab)a^{-1}b^{-1}=a(ba^{-1})b^{-1}=a(a^{-1}b)b^{-1}=(aa^{-1})(bb^{-1})=1\cdot1=1$

The $a^{-1}b^{-1}(ab)$ case is similar.

Derek Luna
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  • Thank you for showing another approach. – Binomi Jan 13 '21 at 22:40
  • You're welcome. I would have used from the start $(ba)a^{-1}b^{-b}$ but then you still have to use more steps with parentheses where you collapse the identity in between elements. – Derek Luna Jan 13 '21 at 22:41
  • Okay but why would you have used that to start? – Binomi Jan 13 '21 at 22:48
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    I meant I could have started $(ab)a^{-1}b^{-1} = (ba)a^{-1}b^{-1}$ instead of what I did, but there are actually atleast the same number of steps. – Derek Luna Jan 13 '21 at 22:52
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$$(ab)^{-1}(ab)=1$$ $$ (ab)^{-1}abb^{-1}=b^{-1}$$ $$(ab)^{-1}a=b^{-1}$$ $$(ab)^{-1}aa^{-1}=b^{-1}a^{-1}$$ $$(ab)^{-1}=b^{-1}a^{-1}.$$

Unknown
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