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MATH 545 HOMEWORK # 1 SOLUTIONS
Here are selected solutions to your first homework assignment. Enjoy!
2.4 (a) closed, (b) not closed, notice 8 + 12 = 20 = 5 mod 15, (c) closed, (d) not closed,
notice 5 · 7 = 35 = 8 mod 9.
2.7 There is no (additive) identity, and it is not closed under addition (the sum of two
odd numbers is not odd).
2.14 (ab)3 = ababab, and (ab−2 c)−2 = (ab−2 cab−2 c)−1 = c−1 b2 a−1 c−1 b2 a−1 .
2.25 Suppose that a and b are arbitrary elements of G. Suppose (ab)−1 = a−1 b−1 . Notice
that a−1 b−1 = (ba)−1 , so (ab)−1 = (ba)−1 , so by inverting both sides, we see that
ab = ba.
2.47 Suppose that x2 = y 2 = e for arbitrary elements x, y ∈ G. Then
xy = xey = x(xy)2 y = xxyxyy = x2 yxy 2 = yx.
3.2 The point of this question is to observe the difference between group operations
within (roughly) the same set, and to recall the definition of a cyclic group. In Q,
the group
1
1
3
1
h i = {k( )|k ∈ Z} = {0, ± , ±1, ± , ±2, . . .}
2
2
2
2
In Q∗ ,
1
1
1
1
1
h i = {( )k |k ∈ Z} = {1, , 2, , 4, , 8, . . .}.
2
2
2
4
8
3.11 Suppose x ∈ R∗ has finite order, say, xn = 1. Notice then that |xn | = |x|n = 1,
and that the only numbers that could qualify are x = ±1. Indeed, these have finite
order of 1 and 2.
3.28 Since H contains 18 and 30, it must contain 30 − 18 = 12. Since H contains 30
and 40, it must contain 40 − 30 − 10. Since H is now known to contain 12 and 10,
H must also contain 2, so it must contain < 2 >= {0, ±2, ±4, ±6, . . .}, the set of
all even integers. If H were to contain anything else, then this quantity would be
odd, and the difference between it and the next smallest number (which is even,
and in H), which would mean that 1 ∈ H, but then H = Z, and it was given that
H is a proper subgroup of Z, and so H must not contain any other elements than
just < 2 >. Therefore, H =< 2 >.
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