My question is:

Let $i = \sqrt{-1}$. Prove that the four roots of unity $\{1, -1, i, -i\}$ form an abelian multiplicative group.

I know that abelian group is a group with commutative property. But I'm not sure how to prove this one. I tried searching through few websites like Wikipedia, YouTube and etc, but couldn't find anything helpful. Can someone please help?

Please forgive my English and structure of the question, I'm new here! :)

  • $\begingroup$ What is it you are having trouble proving? You just need to check that the properties hold for these four elements. $\endgroup$ Feb 10, 2016 at 12:54
  • $\begingroup$ Did you, ever in your life, prove that any set with a given operation is a group? I bet you did. Now simply follow the same steps. $\endgroup$
    – 5xum
    Feb 10, 2016 at 12:55
  • 2
    $\begingroup$ Also, defining $i$ as $\sqrt{-1}$ is nonsense. $\sqrt{x}$ is only defined for $\mathbb{R}^+$. You should define $i$ as one of the solutions of $i^2=-1$. $\endgroup$ Feb 10, 2016 at 13:02
  • $\begingroup$ Maybe it is a stupid question. But I'm having hard time making the multiplication table for this. What should I add for the axis of the table? Should I simply add the element as is and multiply with each other or should I take the fourth root or something? $\endgroup$ Feb 10, 2016 at 13:07
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    $\begingroup$ @TharinduKuruppu see emathzone.com/tutorials/group-theory/group-tables.html $\endgroup$
    – godonichia
    Feb 10, 2016 at 13:15

3 Answers 3


You need to show that the product of two elements of the set $\{\pm1,\pm i\}$ is again an element of the set (*). You need to show that the reciprocal of an element of the set is again an element of the set. You don’t need to check commutativity or associativity, because you’re dealing with complex numbers under multiplication, and those conditions are always satisfied.

(*) The “two elements” above may be equal, too.


Write out the multiplication table for this set and then use it to verify that every group axiom is satisfied.

  • $\begingroup$ Note that this approach could take ages since associativity is not straightforward from the table. $\endgroup$
    – pjs36
    Feb 10, 2016 at 13:34
  • $\begingroup$ I suppose I was assuming familiarity with $\mathbb{C}$ as a group, thereby implying that the associativity axiom is satisfied. $\endgroup$
    – Oiler
    Feb 10, 2016 at 17:47

More generally, if $G$ is an abelian group, then $T_n = \{ g \in G : g^n=1 \}$ is a subgroup of $G$.

Indeed, if $G$ is abelian, then $x \mapsto x^n$ is an endomorphism of $G$ whose kernel is $T_n$.

In your case, $G=\mathbb C^\times$ and $n=4$.


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