Algebra Solutions: 2020

Deduce some properties of a group from its presentation

Let $Y = \langle a, b \ |\ a^4 = b^3 = 1, ab = b^2 a^2 \rangle$ .

Show that $b^2 = b^{-1}$ .
Show that $b$ commutes with $a^3$ .
Show that $b$ commutes with $a$ .
Show that $ab = 1$ .
Show that $a = 1$ and $b = 1$ , hence $Y = 1$ .

From the relation $b^3 = 1$ it follows that $b^{-1} = b^2$ .
Note first that

$b^2 a^3 b$ = $(bbaa)(ab)$

= $(ab)(bbaa)$

= $(a)(b^3)(aa)$

= $a^3$

So that, left multiplying by $b$ , we have $a^3 b = b a^3$ .
Note that $a^9 = a^4 a^4 a = a$ . Hence $a b = a^9 b = (a^3)^3 b = b (a^3)^3 = ba$ , using part 2.
We have $ba = ab = b^2 a^2$ , so that $ba = 1$ . Thus $ab = 1$ .
Since $u^4 = v^3 = 1$ , we have $u^4 v^3 = 1$ . Thus $u (u^3 v^3) = u = 1$ , so that $v = 1$ . Hence $Y = 1$ .

Exhibit an alternate presentation for Dih(4)

Show that $D_4 = \langle x, y \ |\ x^2 = y^2 = (xy)^2 = 1 \rangle$ using the substitutions $x = r$ and $y = s$ .

It suffices to show that $(xy)^2 = 1$ implies $rs = sr^{-1}$ and vice versa. For the forward direction, we have $(rs)^2 = rsrs = 1$ so that $rs = s^{-1} r^{-1} = sr^{-1}$ . For the backward direction, we have $rs = sr^{-1}$ so that $rsrs^{-1} = (rs)^2 = (ab)^2 = 1$ .

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 $y$ .

$b^2 a^3 b$	=	$(bbaa)(ab)$
	=	$(ab)(bbaa)$
	=	$(a)(b^3)(aa)$
	=	$a^3$