Solutions to Problem Set 19

Math 345/504

4-2-2018

1. Let H be the subset of $GL(2,\real)$ consisting of all matrices of the form

$$\left[\matrix{x & 0 \cr 0 & x \cr}\right], \quad\hbox{where}\quad x \ne 0.$$

(a) Prove that H is a subgroup of $GL(2, \real)$ .

(b) Prove that H is normal.

(a) If $\displaystyle
   \left[\matrix{x & 0 \cr 0 & x \cr}\right], \left[\matrix{y & 0 \cr 0
   & y \cr}\right] \in H$ , then

$$\left[\matrix{x & 0 \cr 0 & x \cr}\right] \left[\matrix{y & 0 \cr 0 & y \cr}\right] = \left[\matrix{xy & 0 \cr 0 & xy \cr}\right] \in H.$$

Thus, H is closed under multiplication.

$\displaystyle \left[\matrix{1 &
   0 \cr 0 & 1 \cr}\right] \in H$ : The identity is in H.

If $\displaystyle \left[\matrix{x
   & 0 \cr 0 & x \cr}\right] \in H$ , then

$$\left[\matrix{x & 0 \cr 0 & x \cr}\right]^{-1} = \left[\matrix{x^{-1} & 0 \cr 0 & x^{-1} \cr}\right] \in H.$$

Thus, H is closed under taking inverses.

Hence, H is a subgroup of $GL(2,\real)$ .

(b) Let $\displaystyle
   \left[\matrix{x & 0 \cr 0 & x \cr}\right] \in H$ and $\displaystyle \left[\matrix{a & b \cr c & d \cr}\right] \in
   GL(2,\real)$ . Then

$$\left[\matrix{a & b \cr c & d \cr}\right] \left[\matrix{x & 0 \cr 0 & x \cr}\right] \left[\matrix{a & b \cr c & d \cr}\right]^{-1} = \left[\matrix{a & b \cr c & d \cr}\right] \left[\matrix{x & 0 \cr 0 & x \cr}\right] \left(\dfrac{1}{a d - b c}\right) \left[\matrix{d & -b \cr -c & a \cr}\right] = \left[\matrix{x & 0 \cr 0 & x \cr}\right] \in H.$$

Therefore, H is normal.


2. Use the Universal Property of the Quotient to show that thefunction $f: \integer \to
   \dfrac{\integer}{8 \integer}$ defined by $f(x) = 2 x + 8
   \integer$ induces a function $\tilde{f}: \dfrac{\integer}{4
   \integer} \to \dfrac{\integer}{8 \integer}$ defined by

$$\tilde{f}(x + 4 \integer) = 2 x + 8 \integer.$$

First,

$$f(x + y) = 2(x + y) + 8 \integer = 2 x + 2 y + 8 \integer = (2 x + 8 \integer) + (2 y + 8 \integer) = f(x) + f(y).$$

Hence, f is a group map.

Next, if $4 n \in 4 \integer$ , then

$$f(4 n) = 2(4 n) + 8 \integer = 8 n + 8 \integer = 8 \integer.$$

By the Universal Property of the Quotient, f induces a function $\tilde{f}: \dfrac{\integer}{4
   \integer} \to \dfrac{\integer}{8 \integer}$ defined by

$$\tilde{f}(x + 4 \integer) = 2 x + 8 \integer.\quad\halmos$$


3. Show that the function $f:
   \integer \to \dfrac{\integer}{8 \integer}$ defined by $f(x)
   = 3 x + 8 \integer$ does not induce a function $\tilde{f}:
   \dfrac{\integer}{4 \integer} \to \dfrac{\integer}{8 \integer}$ defined by

$$\tilde{f}(x + 4 \integer) = 3 x + 8 \integer.$$

Specifically, show that f is a group map, but $\tilde{f}$ is not.

First,

$$f(x + y) = 3(x + y) + 8 \integer = 3 x + 3 y + 8 \integer = (3 x + 8 \integer) + (3 y + 8 \integer) = f(x) + f(y).$$

Hence, f is a group map.

However,

$$\tilde{f}[(1 + 4 \integer) + (3 + 4 \integer)] = \tilde{f}(4 + 4 \integer) = \tilde{f}(0 + 4 \integer) = 3 \cdot 0 + 8 \integer = 8 \integer,$$

$$\tilde{f}(1 + 4 \integer) + \tilde{f}(3 + 4 \integer) = (3 + 8 \integer) + (9 + 8 \integer) = 12 + 8 \integer = 4 + 8 \integer.$$

Thus, $\tilde{f}[(1 + 4
   \integer) + (3 + 4 \integer)] \ne \tilde{f}(1 + 4 \integer) +
   \tilde{f}(3 + 4 \integer)$ , so $\tilde{f}$ is not a group map.


What lies in our power to do, it lies in our power not to do. - Aristotle


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