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

Let $n = 2k$ be even with $n \geq 4$ . Show that $z = r^k$ is an element of order $2$ in $D_{2n}$ which commutes with all elements of $D_{2n}$ . Show also that $z$ is the only nonidentity element of $D_{2n}$ which commutes with all of $D_{2n}$ . [cf. §1.1 #33.]

Note that $r^k \cdot r^k = r^{2k} = 1$ , so that $|r^k| = 2$ and in particular, $r^{-k} = r^k$ . Now let $x = s^i r^j$ , $i = 0,1$ and $0 \leq j < n$ be arbitrary in $D_{2n}$ . Then if $i = 0$ we have $r^k \cdot x = r^k \cdot r^j = r^{k+j} = r^j \cdot r^k = x \cdot r^k$ . If $i = 1$ we have $r^k \cdot x = r^k \cdot s r^j = s r^{-k} r^j = sr^kr^j = sr^jr^k = x r^k$ . Thus $r^k$ commutes with every element of $D_{2n}$ .
Now let $x$ be an element of $D_{2n}$ which commutes with every element of $D_{2n}$ . Suppose $x = s r^a$ . Then we have $s r^a \cdot r = s r^{a+1}$ and $r \cdot s r^a = s r^{a-1}$ . Then we have $a+1 = a-1$ mod n, hence $1 = -1$ mod n. But since $n \geq 4$ , this is a contradiction. So $x$ is not of the form $s r^a$ . Now suppose $x = r^a$ . Then we have $s r^a = r^a s = s r^{-a}$ , and thus $a = -a$ mod n, so $2a = 0$ mod n. Thus either $a = 0$ , so that $x = 1$ , or $2a = nq$ for some positive integer $q$ . By a lemma to a previous exercise, since $0 \leq a < n$ we have $2a = n$ and thus $a = k$ . Hence if $n = 2k \geq 4$ is even, the only elements of $D_{2n}$ which commute with all of $D_{2n}$ are $1$ and $r^k$ . $\blacksquare$

Algebra Solutions

Find all the elements in Dih(2n), n even, that commute with all other elements

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