Solutions to Problem Set 15

Math 345/504

3-7-2018

1. (a) Find the order of $(20,
   44)$ in $\integer_{24} \times \integer_{56}$ .

(b) List the elements of the cyclic subgroup of $\integer_8 \times \integer_4$ generated by $(6,2)$ .

(c) Compute the product $(4, (2\
   3))\cdot (6, (1\ 2\ 3))$ in the group $\integer_{10} \times S_3$ . (Multiply permutations right to left, and write the result in disjoint cycle form.)

(a) The order of 20 in $\integer_{24}$ is $\dfrac{24}{(20,24)} = 6$ . The order of 44 in $\integer_{56}$ is $\dfrac{56}{(44,56)} = 14$ . Therefore, the order of $(20,
   44)$ in $\integer_{24} \times \integer_{56}$ is $[6,14] = 42$ .

(b)

$$\langle (6,2) \rangle = \{(0,0), (6,2), (4,0), (2,2)\}.\quad\halmos$$

(c)

$$(4, (2\ 3))\cdot (6, (1\ 2\ 3)) = (4 + 6, (2\ 3)(1\ 2\ 3)) = (0, (1\ 3)).\quad\halmos$$


2. List the elements of the cyclic subgroup $\langle (13, 3) \rangle$ in $U_{14} \times U_{16}$ .

$$\langle (13, 3) \rangle = \{(1, 1), (13, 3), (1, 9), (13, 11)\}.\quad\halmos$$


3. Find a specific element of order 10 in $\integer_8 \times \integer_{35}$ .

Note that $10 = 2 \cdot 5$ . I'll find an element of order 2 in $\integer_8$ and an element of order 5 in $\integer_{35}$ .

There one element of order 2 in $\integer_8$ , namely 4.

There are several elements of order 5 in $\integer_{35}$ ; probably the simplest is 7.

Hence, $(4, 7)$ is an element of order 10 in $\integer_8 \times \integer_{35}$ .


4. $\integer$ is a group under addition. Define $f: \integer \times \integer \to
   \integer$ by

$$f(x, y) = x + y.$$

(a) Prove that f is a group map.

(b) Prove that

$$\ker f = \{(n, -n) \mid \in \integer\}.$$

(a)

$$f[(a, b) + (c, d)] = f(a + c, b + d) = (a + c) + (b + d) = (a + b) + (c + d) = f(a, b) + f(c, d).\quad\halmos$$

(b)

$$f(n, -n) = n + (-n) = 0.$$

Therefore, $(n, -n) \in \ker f$ .

Suppose $(a, b) \in \ker f$ . Then

$$\eqalign{ f(a, b) & = 0 \cr a + b & = 0 \cr b & = -a \cr}$$

Therefore, $(a, b) = (a, -a)$ .

This proves that $\ker f = \{(n,
   -n) \mid \in \integer\}$ .


[Math 504]

5. Find an element of $\integer_6 \times \integer_{10}$ having the largest possible order. Explain why no element can have larger order.

If $(a, b) \in \integer_6 \times
   \integer_{10}$ , then $30 \cdot (a, b) = (30 a, 30 b) =
   (0,0)$ . So every element of $\integer_6 \times
   \integer_{10}$ has order less than or equal to 30.

The element $(1,1)$ has order 30 in $\integer_6 \times \integer_{10}$ . Therefore, it's an element of the largest possible order.


6. Consider the following subset of $\integer_3 \times \integer_6$ :

$$H = \{(0, 0), (1, 1), (2, 2)\}.$$

Either prove that H is a subgroup of $\integer_3 \times \integer_6$ , or give a specific counterexample to one of the subgroup axioms.

H is not a subgroup.

H is not closed: $(1, 1), (2, 2)
   \in H$ , but

$$(1, 1) + (2, 2) = (0, 3) \notin H.$$

Alternatively, $(2, 2) \in H$ , but $-(2, 2) = (1, 4) \notin H$ , contradicting closure under inverses.


The way of progress is neither swift nor easy. - Marie Curie


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