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I read that in order to prove it using groups, we can see that the order of the group of units is equal to Euler's totient function: $$|U(n)|= \phi (n)$$ After this, however, the author goes on to say that this implies that $a^{\phi (n)}=1$ for all $a \in U(n)$. The only reason I can make this connection is that $U(n)$ might be cyclic, in which $a$ to the power of the order of the group is the identity. But what if $U(n)$ is not cyclic? How can we make this logical conclusion then?

GuPe
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2 Answers2

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This is a consequence of Lagrange's theorem.

For $g\in G$, where $G$ is a group, order of $g$ divides $|G|$. This is because the size of the subgroup generated by $g$ must divide $|G|$, and that size must equal the order of $g$.

Therefore, for some integer $k$,

$$ g^{|G|}=g^{o(g)*k}=1^k=1 $$

Apply this result to group of units $U(n)$ with group operation multiplication.

user160738
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  • So $U(n)$ does not need to be cyclic? – GuPe Nov 20 '15 at 01:34
  • No, as long as it is a group of units, then the result will always be true – user160738 Nov 20 '15 at 01:34
  • Just another question, what is the proof that for any finite group $G$, if $g \in G$ there will be a an integer $m$ s.t. $g^m=1$? (Even if $G$ is not cyclic? – GuPe Nov 20 '15 at 01:47
  • If you are asking about Lagrange's theorem, then you might want to look at its wiki page (for instance). As stated, answer to your question is: since $G$ is finite, the list $1,g,g^2,g^3,g^4,...$ must all belong to $G$ (by definition of group) and so the list must have repetition, say $g^r=g^s$ for some $r,s$ integers. Then $g^{r-s}=1$. (By the way, for integers I assume you meant natural number, since $g^0$ is an obvious answer if not. We may assume $r>s$ WLOG here) – user160738 Nov 20 '15 at 01:49
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You don't need Lagrange's theorem when $G$ is a finite abelian group:

If $G$ is a finite abelian group of order $n$ and $a\in G$, then $a^n=1$.

Indeed, consider the map $x \mapsto ax$. This map is injective because $G$ is a group and so is surjective because $G$ is finite. Thus, the map permutes the elements of $G$. Therefore, if the elements of $G$ are elements $a_1, \dots, a_n$, then $a_1 \cdots a_n = (a a_1) \cdots (a a_n) = a^n a_1 \cdots a_n$ and so $a^n=1$.

lhf
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