Show that number of solutions satisfying $x^5=e$ is a multiple of 4?

Solution 1:

Let $G$ be a finite group. Suppose that $a$ and $b$ are two non-identity solutions of $x^5 = e$ in $G$. Then each element of the subgroup $\langle a\rangle$ is also a solution, as is each element of $\langle b\rangle$. But, either $\langle a\rangle = \langle b\rangle$ or $\langle a\rangle\cap\langle b\rangle = e$, since $5$ is prime. Thus, each cyclic subgroup of order $5$ contributes exactly four non-identity solutions, and distinct cyclic subgroups of order $5$ have only trivial intersection, so there is no double counting of the non-trivial solutions from each subgroup. Hence, the number of solutions of $x^5 = e$ is $4$ times the number of distinct cyclic subgroups of order $5$.

Solution 2:

If $x^k=x^j$ and $1\leq j<k\leq4$ then $x^{k-j}=e$ and $1\leq k-j\leq3$. Since $x\neq e$, $k-j$ is 2 or 3. This implies that $x^6=e$. Since $x^5=e$, we have that $x=e$, a contradiction.

Solution 3:

The reasoning of @James can be generalized to any prime $p$. Let $m_p(G):=|\{g \in G : o(g)=p\}|$, then $m_p(G) \equiv0$ (mod $(p-1)$).

But even more is true. The famous proof of James McKay of the Theorem of Cauchy (where he uses a cyclic group of order $p$ acting on the set $\{(x_1, \cdots , x_p) \in G^p: x_1\cdot x_2 \cdots x_p=1\}$) gives as a bonus that $m_p(G) \equiv -1 $(mod $p$).

Combining the two gives $m_p(G)\equiv p-1$ mod $(p(p-1))$.

Solution 4:

We're looking for non-identity solutions of $x^5 = e$, meaning $x \neq e$

$x = x^2$ implies $e = x$ but $x \neq e$
So $x \neq x^2$

$x = x^3 $ implies $x^3 = x^5$ implies $x = x^3 = x^5 = e$ but $x \neq e$
So $x \neq x^3$

$x = x^4$ implies $x^2 = x^5$ implies $x^2 = e$ implies $x^3 = x$ but $x \neq x^3$
So $x \neq x^4$

Is this from an abstract algebra text? I found the same question, but I think there's a typo- my book says "multiple of five". I could only show multiples of four.