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I can't seem to find anything on the topic.

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    $\begingroup$ yes, itself !!!! $\endgroup$ – Anubhav Mukherjee Mar 19 '16 at 13:08
  • $\begingroup$ Think it out. Let $g = e^{-1}$. Then $g*e = e^{-1}*e = e$. But $g*e = g$. So $g = e$. End of story. $\endgroup$ – fleablood Mar 19 '16 at 16:41
  • $\begingroup$ ... well, not quite end of story. Let $g = e^{-1}$. Then $g*h = e*(g*h) = (e*g)*h = e*h = h = h*e = h*(g*e) = (h*g)*e = h*g$. So for all $h$, $g*h = h* g = h$ so $g$ (whatever it is) is an identity element. No group has two identity elements; because then $g*e = e$ and $g*e = g$ which is a contradiction. So $g = e$. $\endgroup$ – fleablood Mar 19 '16 at 16:47
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All elements of a group have an inverse. This is a requirement in the definition of a group.

For an element $g$ in a group $G$, an inverse of $g$ is an element $b$ such that $gb = e$ where $e$ is the identity in the group. (Since the inverse of an element is unique, we usually denoted the inverse of $g$ $g^{-1}$ or $-g$.)

Note now that $ee =e$, so by definition $e$ is an inverse of itself.

You might be wondering if other elements might be their own inverses. The answer to this is yes. For example in the group $\mathbb{Z}_2$ of order $2$, both elements are their own inverse.

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    $\begingroup$ In addition, all elements in $\mathbb{Z}_2 \times \mathbb{Z}_2 \times ... \mathbb{Z}_2$ are their own inverse, and thus there is no bound on the size of such groups. $\endgroup$ – Nate 8 Mar 20 '16 at 3:46
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Let $1$ be the identity element. Then its inverse $a$ is defined by $$ 1\cdot a = 1 $$ But $1 \cdot x = x$ so in particular $1 \cdot a = a$. Therefore, $a = 1$ and $1$ is its own inverse.

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If $e$ is the neutral element of $G$ then $eg=ge=g$ for all $g\in G$, in particular for $g=e$ then $ee=e$, ie $e=e^{-1}$.

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Let $g = e^{-1}$.

Then $g*e = e*g = e$. Solve for $g$.

Then $g*e = e \implies g*e*e^{-1} = e*e^{-1} => g = e$.

This gets kind of silly the more you think about it.

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  • $\begingroup$ Nice proof. You used only the inverse axiom. $\endgroup$ – john Oct 5 '17 at 16:52
  • $\begingroup$ Gad, I can't remember writing this. I think I was trying to be sarcastic. Or maybe I was trying to be helpful. As $e*e = e$ it should be clear $e = e^{-1}$ and maybe I was saying "if you don't get it... then solve it". The solve $x*g = h$ you do $x= x*g*g^{-1} = h*g^{-1}$ so to solve $xe = e$ you do $x=x*e*e^{-1} = e*e^{-1}=e$. Most text take this as self-evident. I'm not sure if I was trying to say "look you dummy" or "see, this actually makes perfect sense". I really don't remember. $\endgroup$ – fleablood Oct 5 '17 at 17:09

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