Does someone know the why raising the element of a group to the power of the order of the group yields the identity?

By (finite) group I mean a tuple (G,*) that satisfies the following:

  • closure under operation *
  • associative under operation *
  • has identity element
  • has inverse

I was unsure why:

$$ a^{|G|} = I $$

[I am aware that its a basic fact but I was unable to find the proof online [don't know the name of the theorem]). I wanted to try to prove it myself, but I was not sure how to because, the statement of the theorem does not seem to state anything about what is allowed to be an element of the group and what isn't allowed, does it only apply to integers? What is allowed to be an element of the group (binary strings?)? Also, I was not sure what a raised to the order meant, because, the group might not have "multiplication" defined, maybe it has its own way of doing stuff to its elements, so I didn't know what raising it to the power of meant. Does it mean using operation * |G| times? Does this only apply to cyclic groups?

I usually post my (failed) attempts to solve the question first but I was not sure how to start proving anything because I was not sure what the elements of the group were allowed to be or what the raising meant with respect to operations on the elements of the group.

Can anyone state the name of the theorem?

  • 3
    $\begingroup$ Once you have Lagrange's theorem, this follows trivially. So keep on reading whatever book you are using to learn group theory, and you will see it soon. $\endgroup$ – fkraiem Jan 7 '14 at 2:10
  • $\begingroup$ do you recommend any free book I could use online? $\endgroup$ – Pinocchio Jan 7 '14 at 2:18
  • $\begingroup$ No, sadly the only online resources are know at that level are some lecture notes which are in French, but others might know some. Feel free to ask another question about this with the reference-request tag. $\endgroup$ – fkraiem Jan 7 '14 at 2:23
  • $\begingroup$ the link to my reference question for that is here: math.stackexchange.com/questions/629759/… $\endgroup$ – Pinocchio Jan 7 '14 at 3:07

If $G$ has $|G|$ many elements, then the set $$\{a, a^2, a^3, \cdots, a^{|G|}\}$$

either has a repetition or exhaust the group. If it exhaust the group, one of these elements is $1$. If there is a repetition, so $a^{r} = a^{s}$ for $r>s$, then $a^{r-s} = 1$. In any event, at least one of these elements must equal $1$, say $a^n = 1$, and choose $n$ to be the smallest such exponent. If we can show that $n$ divides $|G|$, we are done. But $n$ is the order of the subgroup $$\{a, a^2, \cdots , a^n(=1)\}$$

And by Lagrange's theorem, the order of a subgroup must divide the order of the group.

To rephrase here a bit: If $a$ is a generator of $G$, then raising it to the power of the order of the group guarantees that it will cycle through all the elements and return to the identity. If $a$ is not a generator of $G$ then Lagrange's theorem guarantees that the order of the subgroup generated by $a$ divides $|G|$ and therefore if $a^{|a|}=1$ then $(a^{|a|})^{u}=1$ where $|G|=u|a|$.

In response to your questions about definitions: A group is an abstract object. We do not know anything about the sorts of objects inside the group. All we know about the group is that it satisfies certain axioms and has a (binary) group operation $\ast$. Given a pair of elements $a,b \in G$, we write $a \ast b$ or $ab$ to denote the group operation acting on the pair of elements. The exponent notation means: $a^{2} = aa, a^{7} = aaaaaaa$.

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    $\begingroup$ It is not true that the powers of $a$ exhaust $G$, unless $G$ is cyclic and $a$ is a generator. $\endgroup$ – fkraiem Jan 7 '14 at 2:16
  • $\begingroup$ Your initial answer and the remark fkraiem was on doubt I had or something I was not sure if I understood from your answer. Does that mean that, every group is cyclic? I am sure that must not be the case, not sure how to determine if something is cyclic but if you have any clarifying comment it would be awesome! :) $\endgroup$ – Pinocchio Jan 18 '14 at 6:05
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    $\begingroup$ I am sorry if this is trivially obvious for you but, why are we done if n divides $|G|$? $\endgroup$ – Pinocchio Jan 18 '14 at 6:07
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    $\begingroup$ Oh I see, if n divides $|G|$ then $|G|=nk$ for some k. Thus, if we define n such that $x^{n} = 1$ then $x^{|G|}=x^{nk}=(x^{n})^{k} = (1)^{k} = 1$ which is true from Lagranges theorem. However, the only part I dont buy yet (or dont understand) just to be 100% convinced, why does n have to necessarily be the order of the subgroup (I understand that if it is true, then it follows the fact I wrote but how come that fact is even true)? i.e. If $H \subset G$ What is the proof that $x^{|H|}=1$? Thnx for the help btw! Why is raising to the power of the subgroup give 1? :D $\endgroup$ – Pinocchio Jan 18 '14 at 6:52
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    $\begingroup$ Also, doesn't the proof for $a^{|G|} = 1$ need that there exists a subgroup of $G$ that is cyclic for us to apply the fact that $a^{k|S|} = 1$? So every group has a cyclic subgroup? $\endgroup$ – Pinocchio Jan 18 '14 at 18:33

Because of Lagrange's theorem, the order of a subgroup divides the order of the group. Now consider the cyclic subgroup generated by a.

  • $\begingroup$ Doesn't the argument require that there exists a subgroup of $G$ that is cyclic for us to apply the fact that $a^{k|S|} = 1$ (by Lagrange's thm)? Thus, for any group does that mean that there exists a cyclic subgroup? $\endgroup$ – Pinocchio Jan 18 '14 at 18:34

Here's a way to get the result without passing through Lagrange's theorem.

Let $g\in G$, and take its powers until we get to the identity again (which must happen if the group is finite). We thus get a sequence $\langle g\rangle\equiv\{g,g^2,...,g^{n-1},e\}$. If $|G|=n$, then we are done, so let's assume that this is not the case, and thus $n<|G|$.

There must then be some $h\notin\langle g\rangle$. If we multiply each element of $\langle g\rangle$ by $h$, we must necessarily get an entirely different set of $|n|$ elements. Indeed, for any $j,k$, $h g^j=h g^k$ implies $j=k$. It follows that $h\langle g\rangle$ has no intersection with $\langle g\rangle$.

We can now just iterate the same reasoning until we run out of elements of $G$: find a new element $x\in G$ that is not contained in the sequences previously considered. Every time we consider such a new element, we count $n$ brand new elements of $G$. At some point, we must necessarily run out of new elements of $G$ to choose from, but because we counted them $n$ at a time, it follows that $|G|$ must be a multiple of $n$: $|G|=kn$ for some integer $k$. But if $g^n=e$, then clearly $g^{nk}=g^{|G|}=e$.


Let $|G| = n$. Take any element $x \in G$ and consider the smallest positive integer $p$ such that $x^p = 1$. By definition, $|x| = p$. The order of any cyclic group is equal to the order of its generator. In particular, this is true for the cyclic group generated by $a$, that is, $|\langle a\rangle| = p$. Since $\langle a\rangle$ is a subgroup of $G$ (see below), by Lagrange's theorem, $n = pq$ for some integer $q$. Then:

$$x^{n} = x^{pq} = (x^p)^q = 1^q = 1$$

So $x^n = 1$ for any $x \in G$.

About your question:

doesn't the proof for $a|G|=1$ need that there exists a subgroup of G that is cyclic?

Yes. And such a group does exist. In fact, there are many. Take any element $a \in G$ and consider, as Elchanan mentioned, the group generated by $a$: $ \langle a\rangle = \{a, a^2, \ldots a^{|G|}\}$. This is, by definition, a cyclic group. And it is a subgroup because it is closed under multiplication: ($a^ia^j = a^{i + j}$); closed under inverses: $(a^i)^{-1} = (a^{-1})^i$; and, finally, $a^{|G|} = 1$.

  • $\begingroup$ For the statement before last, did you mean inv(a^i) = inv(a)^(-i) ? Examples: inv(4+3 mod 5) = inv(4)^(-3) mod 5 => (3, 3); inv(1+3 mod 5) = inv(1)^(-3) mod 5 => (1, 1); inv(2+3 mod 5) = inv(2)^(-3) mod 5 => (0, 0) $\endgroup$ – גלעד ברקן Feb 27 '19 at 11:24
  • $\begingroup$ Also, I don't think there is a guarantee that an arbitrary element, a, of G will generate a group of order |G|, as you seem to imply. That would mean the enumeration for the set generated by a should end with a^n, not a^|G|; and the last statement should be a^n = 1 for some n. $\endgroup$ – גלעד ברקן Feb 27 '19 at 12:46

Let $G$ be an abelian group. Let $a\in G$. Denote $aG=\{ag|g\in G\}$. Then $aG=G$ since $a$ is invertible.

Multiplying the elements together gives $$a^{|G|}\prod_{g\in G}g=\prod_{g\in G}g.$$

Cancelling now gives $$a^{|G|}=1.$$

(Compare this with Fermat's Little Theorem)

Can you see that this argument needs $G$ to be abelian? You can almost assume $G$ is abelian by just considering the subgroup generated by $a$ but then you only get that some power of $a$ is the identity rather than $|G|$. Knowing that this power divides $|G|$ would finish the argument but at this point the answer is essentially the same as previous answers.


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