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This question already has an answer here:

In chapter 11 of Contemporary Abstract Algebra I learned Lagrange theorem:

If $G$ is a finite group and $H$ is a subgroup of G, then $|H|$ divides $|G|$

and one of its corollaries:

In a finite group, the order of each element of the group divides the order of the group

In the exercise I've been asked to prove, using that corollary, that the order of the multiplicative group of integers modulo $n$ is even when $n>2$. Or in other words, that the Euler $\phi(n)$ is even for all $n>2$.

I've seen some proofs of that claim but non are based on that corollary. How can I prove that using the corollary above?

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marked as duplicate by Tobias Kildetoft, Henning Makholm, DonAntonio group-theory Sep 28 '16 at 10:51

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    $\begingroup$ Actually, the answer I gave in the link you mention precisely uses that result. $\endgroup$ – Tobias Kildetoft Sep 28 '16 at 10:36
  • $\begingroup$ You wrote "odd" in your question's title, and then "even". Decide which one it is. $\endgroup$ – DonAntonio Sep 28 '16 at 10:38
  • $\begingroup$ Your title is kind of "odd". It is not consistent with the body. $\endgroup$ – drhab Sep 28 '16 at 10:38
  • $\begingroup$ sorry - fixed it $\endgroup$ – Bush Sep 28 '16 at 10:39
  • $\begingroup$ Saw your answer there, that's what I looked for. Should I close? $\endgroup$ – Bush Sep 28 '16 at 10:51
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If $n>2$ then $\pm 1$ are not congruent mod $n$.

Consider the group $G = (\mathbb{Z}/n\mathbb{Z})^{\times}$ of integers mod $n$. We have an obvious subgroup $H = \{\pm 1\}$ and so by Lagrange we must have $|H| = 2$ divides $|G| = \phi(n)$.

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