Algebra Solutions: Find all the elements in Dih(2n), n odd, which commute with all other elements

Show that if $n \geq 3$ is odd, then 1 is the only element of $D_{2n}$ which commutes with all of $D_{2n}$ .

Recall that every element of $D_{2n}$ is of the form $s^i r^j$ where $i = 0,1$ and $0 \leq j < n$ . Now suppose $x \in D_{2n}$ commutes with all of $D_{2n}$ . If $x = sr^a$ , then we have $r \cdot s r^a = s r^{a-1}$ and $s r^a \cdot r = s r^{a+1}$ , so that $a+1 = a-1$ mod n. Then we have $1 = -1$ mod n, a contradiction since $n \geq 3$ . Thus $x$ is not of the form $sr^a$ . Now suppose $x = r^a$ for some $0 \leq a 0$ . If $q \geq 2$ we have $2a < 2n \leq qn$ , a contradiction. If $q = 1$ , we have $n = 2n$ even, a contradiction.
Thus 1 is the only element of $D_{2n}$ which commutes with all of $D_{2n}$ .

Algebra Solutions

Find all the elements in Dih(2n), n odd, which commute with all other elements

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