My lecture notes say that in a group $G$, the identity $e$ is the only element of $G$ which has order $1$.

I would like to know why the order of $e$ is said to be $1$, when it could clearly be $0$ since the identity $e$ acted on itself zero times is just $e$.

Is it valid to think about an element acting on itself zero times? So for example is it valid to even think about why $e^0 = e$? Cheers.

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    $\begingroup$ If $g$ is any element, then $g^0=e$. So should every $g$ have order zero? $\endgroup$ May 19, 2017 at 5:51
  • $\begingroup$ @LordSharktheUnknown Shouldn't any $g^0 = g$? We are taking an element of the group, $g$, and doing nothing with it. We just get back $g$. $\endgroup$ May 19, 2017 at 5:52
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    $\begingroup$ So then $g=g^1=g^{0+1}=g^0g^1=gg=g^2$??? $\endgroup$ May 19, 2017 at 5:55
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    $\begingroup$ @TristanBatchler Are you familiar with the fact that for any real number $x$, $x^0=1$? 'Raising to the power 0' isn't the same as 'doing nothing' in any part of mathematics. $\endgroup$
    – dbmag9
    May 19, 2017 at 13:27
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    $\begingroup$ @TristanBatchler Using that analogy, $g^0$ would be an 'empty product', rather than anything about $g$ acting on itself. $\endgroup$
    – dbmag9
    May 19, 2017 at 13:39

4 Answers 4


The order of an element is a positive integer, by definition.

  • $\begingroup$ Thanks. This was not in my lecture notes so it got me thinking. I now realise it even says it must be positive on the wikipedia page, which I should have paid more attention to. Sorry! $\endgroup$ May 19, 2017 at 6:14
  • $\begingroup$ To clarify, this would be one of those rules that does not fit into the formula like @TristanBatchler has presented, so we have an addendum for input is identity? $\endgroup$
    – Pysis
    May 19, 2017 at 13:45

Don't think of $g^0$ as representing $g$ acting on itself $0$ times but as $g$ acting on something else $0$ times. Groups frequently represent actions on other objects. An important class of examples is symmetry groups.

Consider G as the group of symmetries of a square and $g \in G$ as rotate clockwise by $90^\circ$. Then $g^2$ is rotate clockwise by $180^\circ$ and $g^3$ is rotate clockwise by $270^\circ$. So, what is a sensible definition of $g^0$ in this context? Well, rotate by $0^\circ$. Similarly we can sensibly define $g^{-1}$ as rotate anticlockwise by $90^\circ$ since this will undo the effect of $g$. With these definitions, $g^n$ is rotate clockwise by $n \times 90^\circ$ whether $n$ is positive, zero, or negative and $g^m \times g^n = g^{m+n}$ and life is good. With your convention, things would not be so neat.

So, with this convention, $g^0$ is always $e$ and hence to be useful we must require the order to be $> 0$ and not just $\geq 0$.


You can define powers of a group element $g$: $g^0=e$ and $g^{n+1}=g^n\cdot g$ for $n\geq 0$. The order of an element is chosen to be $\geq 1$.

  • $\begingroup$ But why define $g^0 = e$? Thinking about it, one would define any $g^0 = g$ because we are taking an element of the group, $g$ and doing nothing with it. We just get back $g$. At least that makes more sense in my mind... $\endgroup$ May 19, 2017 at 5:55
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    $\begingroup$ @Tristan Batchler Why force the issue when in fact the order is defined the least positive integer $n$ for which $g^n=e$, $\endgroup$ May 19, 2017 at 6:05
  • $\begingroup$ Whoops, I guess I missed the part here the order had to be a positive integer. This was not in my lecture notes so it got me thinking. $\endgroup$ May 19, 2017 at 6:13
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    $\begingroup$ @TristanBatchler $g^0$, interpreted literally, is the product of $0$ copies of $g$, not "doing nothing" with $g$. This is an instance of the "empty product", which for various reasons is defined to be the identity element. Here is one argument: We should be able to replace $g^n$ with $n$ copies of $g$. So $hg^n$ should be the same as $hg\dots g$, where $g$ is repeated $n$ times. Setting $n=0$, we get $hg^0 = h$. Similarly, we should have $g^0h = h$. This means $g^0$ is the identity element. This isn't a proof, just a reason for the definition, and it doesn't work for negative values of $n$. $\endgroup$ May 19, 2017 at 6:30

In my book, the order of an element is the order of the cyclic group it generates, so the identity clearly has order 1.


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