# If $a^3=e$, then $a$ has a square root

Assuming $a\in G$ where $G$ is a group.

I'm not sure why this is hard for me. Essentially, the problem is just saying:

If $a^3=e$, then there exists $x \in G$ such that $a=x^2$.

Can somebody give me a hint or a direction to start in? (not full solution, please--I want to figure it out myself)

Thank you.

Hint: Multiply by $a$ to find $$a^4 = a$$

• Okay, so I tried the following: – Patrick Shambayati Aug 21 '14 at 14:28
• $a^4=a$, so $a^4=x^2$, which I re-wrote as $(a^2)(a^2)=(x)(x)$. Is it true that $x=a^2$? (Does this follow the rules of algebra?). If so, what actual steps can I use to show that? I guess I am probably being too cautious with all these new rules I have learned for abstract algebra – Patrick Shambayati Aug 21 '14 at 14:30
• It needn't be unique. But yes, $a^2$ is a square root. – JHance Aug 21 '14 at 14:32
• It's not inherently true since you can't guarantee that $a^2$ is the only solution to $x^2 = a^4$ (think of $-a^2$ in the reals, for example). However, this allows you to show that $a^2$ is a square root of $a$ wich suffices to prove its existence – AlexR Aug 21 '14 at 14:32
• Oh I see. Re-reading the problem, it becomes clear that I am not finding square roots--just proving the existence of at least one. – Patrick Shambayati Aug 21 '14 at 14:37

Hint: Try $x$ as a power of $a$ (since these are the only things being assuredly in $G$).

Hint $\,\ a^{\large 3}= 1\,\Rightarrow\,a^{\large 3n}=1\,\overset{\times\, a}\Rightarrow\,a^{\large 1+3n} = a\,$ is a square if $\ 2\mid 1\!+\!3n,\,$ e.g. $\, n = \,\ldots\$ QED

Remark  Here is the intuition. $\ a^{\large 3}= 1\$ implies that exponents on $\,a\,$ can be considered mod $\,3,\,$

$$\color{#c00}{a^{\large 3}= 1}\ \Rightarrow\ a^{\large k+3n} = a^{\large k} (\color{#c00}{a^{\large 3}})^{\large n} = a^{\large k}\ \ \ {\rm so}\ \ \ a^{\large j}\! = a^{\large j\ {\rm mod}\ 3}$$

Therefore we can replace the exponent $1$ in $\,a^1\,$ by any $\,j\equiv 1\pmod 3,\,$ which includes  even $\,j,\,$ i.e. $\,{\rm mod}\ 3,\,$ we have that $1$ is "even", i.e. $\ 2\mid 1,\,$ i.e. $2$ is invertible. This generalizes as follows.

If $\,\color{#c00}{a^k = 1}\,$ and $\,\gcd(n,k)=1\,$ then $\,a\,$ is an $n$'th power. Indeed, by above it suffices to find a multiple $\,jn\,$ of $\,n\,$ that is $\,\equiv 1\pmod k,\,$ i.e. to invert $\,n\,$ mod $\,k,\,$ which is easy:

$\qquad\qquad$ by Bezout, there are $\,i,j\in\Bbb Z\,$ with $\ jn = 1 + ik\$ so $\ (a^j)^n = a(\color{#c00}{a^k})^i = a$

Note how the problem reduces to the problem of division mod $\,k.\,$ The structure underlying this reduction will become clearer when one studies cyclic groups and modules (over $\,\Bbb Z/k)$.

• The donwvotes are quite puzzling. As always, if something is not clear then please feel welcome to ask questions and I will be happy to elaborate. – Bill Dubuque Aug 21 '14 at 15:57
• I did not downvote, but I suspect the downvotes are for for two reasons. Firstly, the question already has two perfectly good answers. Secondly, it seems that you are over-complicating what is actually a rather simple question. I mean, "$a^4=a$" gets the point across, and the student can work the rest out quite easily... – user1729 Aug 21 '14 at 16:07
• @user1729 In my experience, most students will not grasp the essence of the matter from what little was said in the other answers. There is large pedagogical abyss between most textbooks in elementary number theory and abstract algebra. This often leads to conceptual gaps in matters like this. That's why I often elaborate on such topics - to help students overcome those ubiquitous gaps. – Bill Dubuque Aug 21 '14 at 16:18
• See also here for further elaboration. – Bill Dubuque Oct 11 '18 at 21:01

Heres another way to view it simply:

$$a^3=e$$ $$aaa=e$$ $$(aa)a=e$$ $$(a^2)a=e$$ $$(a^2)a(a^{-1})=e(a^{-1})$$ $$a^2(aa^{-1})=e(a^{-1})$$ $$a^2=a^{-1}$$  

Where if you have defined $a \in G$ will imply $a^{-1} \in G$ by the definition of a group