Prove that a cyclic group can have no more than one element of order two.

Question

(1)Why can't a cyclic group have more than one element of order two?

(2)Why does the group $U(n^2 -1)$ have to have more than one element of order two?

Answer 1

A cyclic group is addition modulo the number of elements. That number has at most one integer that is one half it.

Answer 2

Hints:

1) If $\;G\;$ is cyclic group of order $\;n\;$ , then for any divisor $\;d\;$ of $\;n\;$ there exists one unique subgroup of order $\;d\;$ in $\;G\;$ , which of course is also cyclic.

2) A cyclic group of order $\;n\;$ has $\;\varphi(n)\;$ different generators $\;=\varphi(n)\;$ elements of order $\;n\;$

The above , with $\;\varphi=$ Euler's Totient Function.

Answer 3

This is an answer to your question (2) (as other answers have dealt with (1)). Note that if $n = 2$, then $n^2 - 1 = 3$ and $U(3) \cong \mathbb{Z}/2\mathbb{Z}$ does have a unique element of order $2$, so we must assume $n \ge 3$.

The structure of $U(n)$ depends on the prime factorization of $n$. If $n^2 - 1 = p_1^{e_1}\cdots p_k^{e_k}$ with $p_i$ prime, then $U(n^2-1) \cong U(p_1^{e_1}) \times \cdots \times U(p_k^{e_k})$. We have the following cases:

i) There are two distinct odd $p_i$: for odd primes $p$, $U(p^n)$ contains an element of order $2$, so in this case $U(n^2-1)$ contains a copy of $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, and thus $3$ elements of order $2$.

ii) $n^2-1 = 2^{e_1}p_2^{e_2}$: in this case $n$ is odd, so $n^2-1 \equiv 0 \pmod{4}$, hence $e_1 \ge 2$, and $|U(2^{e_1})| = 2^{e_1-1}$, so again $U(n^2-1)$ contains $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$.

iii) $n^2 - 1 = p^e$: writing $n^2 - 1 = (n+1)(n-1)$, by uniqueness of factorization, $n+1 = p^a$, $n-1 = p^b$ with $a > b$, but then $n+1 \ge p(p^b) = p(n-1) \implies 1 \ge (p-1)n - p$. If $n \ge 4$, this is impossible, and if $n = 3$, this is only possible for $p = 2$, in which case $U(n^2 - 1) = U(8)$ is isomorphic to $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$.

Answer 4

(1) A cyclic group cannot have more than one element of order 2. The number of elements in a cyclic group of order 2 can be only $\phi(2)$ =1.

(ii) $U(n^2 -1)$ refers to the elements prime to $ n^2 -1 $ under modulo multiplication $ n^2 -1 $

HINT : An element of order 2 means it is self inverse. Suppose that there exists only one element of order 2 and proceed