Algebra Solutions: Deduce some properties of a group from its presentation

Let $X_{2n} = \langle x, y \ |\ x^n = y^2 = 1, xy = yx^2 \rangle$ .

Show that if $n = 3k$ , then $X^{2n}$ has order 6, and that it has the same generators and relations as $D_6$ with $x = r$ and $y = s$ .
Show that if $\mathsf{gcd}(3,n) = 1$ , then $x$ satisfies the additional relation $x = 1$ . In this case, deduce that $X_{2n}$ has order 2. (Use the fact that $x^n = x^3 = 1$ .)

Note that, according to the text, every element of $X_{2n}$ can be written as $y^a x^b$ with $0 \leq a < 2$ and $0 \leq b < n$ . Moreover, we have $x = xy^2 = xyy = yx^2y = y^2 x^4 = x^4$ , so that $x^3 = 1$ . Thus $X_{2n} = \langle x, y \ |\ x^3 = y^2 = 1, xy = yx^2 \rangle$ , and in fact we can write elements of $X_{2n}$ as $y^a x^b$ where $0 \leq a < 2$ and $0 \leq b < 3$ , so that $|X_{2n}| \leq 6$ .

If $n = 3k$ , we have $x^n = x^{3k} = 1^k = 1$ . Since $x$ and $y$ are distinct, then, none of the elements $y^a x^b$ with $0 \leq a < 2$ and $0 \leq b < 3$ can be equal. Thus $|X_{2n}| = 6$ . Moreover, if we let $x = r$ and $y = s$ , then $X_{2n}$ clearly satisfies the same relations as $D_6$ .
If $\mathsf{gcd}(3,n) = 1$ , then we have either $n = 3k + 1$ , so that $x = 1$ , or $n = 3k + 2$ , so that $xy = yx^2 = y$ and thus $x = 1$ . In either case, $x = 1$ , so that the elements of $X_{2n}$ are precisely $1$ and $y$ .

Algebra Solutions

Deduce some properties of a group from its presentation

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