M501: Assignment 2

Jason Crothers

September 11, 2015

2.1.6 Let H = {g ∈ G : |g| <∞}. We need to show that H is nonempty and that xy−1 ∈ Hfor all x, y ∈ H. The order of the element 1 ∈ G is 1, so 1 ∈ H 6= ∅.Let x, y ∈ H. There exists positive integers n,m such that

xn = ym = 1

Recall that |y−1| = |y|. Because H is Abelian, we have

(xy−1)nm = xnm(y−1)nm = (xn)m((y−1)m)n = 1m1n = 1

So xy−1 have finite order which means xy−1 ∈ H.

2.1.9 Let A ∈ SLn(F). Then |A| = 1 so A is invertible and A ∈ GLn(F), so SLn(F) ⊆ GLn(F).Now it suffices to show that SLn(F) is nonempty and that AB−1 ∈ SLn(F) for allA,B ∈ SLn(F).

The identity matrix In×n has determinant 1, so SLn(F) is not the empty set. LetA,B ∈ SLn(F). By the properties of determinants

|AB−1| = |A||B−1| = |A||B|

=1

1= 1,

so AB−1 ∈ SLn(F).

2.2.6 (a) We need to show that for any h ∈ H the following holds

hHh−1 = H.

Fix h ∈ H. First we will show that hHh−1 ⊆ H. For any h0 ∈ H, because H isa group

hh0h−1 ∈ H

Thus, hHh−1 ⊆ H.

For the reverse inclusion, for any h0 ∈ H we can write

h0 = 1h01 = hh−1h0hh−1 = h(hh0h

−1)h−1 ∈ hHh−1

This holds because h−1h0h ∈ H. Thus hHh−1 = H for all h ∈ H and H ≤ NG(H).

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Let D2n be the dihedral group of order 2n where n ≥ 3. Let H = {s, r}. This setis not a subgroup, and

srs−1 = srs = r−1 6∈ {s, r}

Therefore H is not a subset of NG(H).

(b) First suppose that H is Abelian, then for any x, y ∈ H

xyx−1 = xx−1y = 1y = y

so H ≤ CG(H).

Suppose now that H ≤ CG(H). If x, y ∈ H, then x = yxy−1 and

xy = (yxy−1)y = yx

Thus, H is Abelian.

2.2.7 (a) Let n ≥ 3 be odd. We will show that s only commutes with itself and 1 whichimplies that Z(D2n) = 1.

Let x ∈ D2n. Then x = rks or x = rk for some integer k. If x = rks then

s(rks) = r−k

and(rks)s = rk

So, if s and rks commute, rk = r−k and r2k = 1. This means n | 2k, but n is oddso n | k which implies rk = 1. Thus, x = s. Suppose x = rk and commutes withs. Then

rks = srk =⇒ srks = rk

But, srks = r−k, so again we get r2k = 1 and by the same argument as aboven | k and x = 1. This completes the proof.

(b) First we show that rk ∈ Z(D2n). Notice that because n = 2k,

(rk)(rk) = r2k = 1,

so rk = r−k. From this srk = rks. We already know that rk commutes with allpowers of r so we need to show this is true for elements rms,

rk(rms) = rmrks = (rms)rk

Thus, rk ∈ Z(D2n).

We will now show that 1 and rk are the only elements of this set. Suppose thatrm ∈ Z(D2n) and rm 6= 1, then rm commutes with s. By the argument in part(a), we know this implies that r2m = 1. Thus, n = 2k | 2m which means k | m.This means rm is in the cyclic group generated by rk. But rk has order 2, sorm = 1 or rm = rk. Since we said rm is not equal to 1, rm = rk.

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Suppose that rms ∈ Z(D2k). By the same argument as above, this implies thatrms = rks. If this element is in Z(D2k), then it commutes with r, so

r(rks) = rk+1s

and(rks)r = srk+1

are equal. But, this implies that r2(k+1) = 1 which means

r2(k+1) = r2kr2 = r2 = 1

which is a contradiction since n > 2. Thus, Z(D2n) = {1, rk}.

2.3.3 The group Z/48Z is generated by the element 1̄. For any x̄ ∈ Z/48Z

|x̄| = |1̄|(|1̄|, x)

=48

(48, x)

The element x̄ will be a generator only if and only if the above equation is equal to48 which is true if and only if (48, x) = 1. There are 16 integers 1 ≤ n < 48 suchthat (48, n) = 1. Thus, there are 16 distinct generators in Z/48Z. Here is a list ofrepresentatives of the congruence classes that generate Z/48Z:

1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47

2.3.8 Since 1̄ generates the group Z/48Z,

ϕa(n̄) = ϕa(1̄ + · · ·+ 1̄)︸ ︷︷ ︸n−times

= ϕa(1̄) · · ·ϕa(1̄)︸ ︷︷ ︸n−times

= (xa)n

So the image of ϕa is the group 〈xa〉. Thus, to make ϕa surjective, we need 〈xa〉 =Z/48Z which is true if and only if (a, 48) = 1. Since both groups are finite, this impliesϕa is injective as well and is an isomorphism.

2.3.12 (a) Because (x, 1) 6= (1, 1) and

(x, 1)2 = (x2, 12) = (1, 1),

(x, 1) has order 2. Also, because (1, x) 6= (1, 1) and

(1, x)2 = (12, x2) = (1, 1),

(1, x) has order 2. Since (1, x) 6= (x, 1), the group Z2×Z2 has 2 distinct subgroupsof order 2. Therefore, Z2 × Z2 is not cyclic.

(b) Suppose Z2 × Z is cyclic. Since Z2 × Z has infinite order, it must be isomorphicto Z. The element (x, 0) has order 2 since (x, 0) 6= (1, 0) and

(x, 0)2 = (x2, 2 · 0) = (1, 0).

This is a contradiction since Z contains no element of order 2.

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(c) Suppose that Z× Z is cyclic, so

Z× Z = 〈(x, y)〉

Then there is an integer n such that

(x, y)n = (nx, ny) = (1, 0)

This means that nx = 1 so n cannot be zero. Also ny = 0, and since n isnonzero, y = 0. This is a contradiction since (x, 0) does not generate whosesecond component is not zero. Thus, Z× Z is not cyclic.

2.3.23 It suffices to show that there are two distinct subgroups of order 2, for if G = (Z/2nZ)×

was cyclic these two subgroups would be equal. The elements 2n − 1 and 2n−1 + 1 areboth odd, so they are members of G. Also,

2n − 1 6≡ 1 mod 2n

and2n−1 + 1 6≡ 1 mod 2n.

Observe that(2n − 1)2 ≡ (−1)2 ≡ 1 mod 2n

And because 2n− 2 ≥ n for all n ≥ 3, we have

(2n−1 + 1)2 ≡ 22n−2 + 2n + 1 ≡ 1 mod 2n

Thus, both of these elements have order 2 which means {1, 2n − 1} and {1, 2n−1 + 1}are distict subgroups of order 2.

2.4.14 (a) Let G be a finite group, so G = {x1, . . . , xn}. Then clearly G = 〈x1, . . . , xn〉, soG is finitely generated.

(b) The group Z is cyclic and is finitely generated by {1}.(c) Suppose that H is a finitely generated subgroup of Q and let{

a1b1, . . . ,

anbn

}be a set of generators of H. Let k = b1 · · · bn. Then for ai

bi,

(aib1 · · · bi−1bi+1 · · · bn)1

k=

aibi

Since every element in the set of generators of H is in 〈 1k〉, it must be that H ≤ 〈 1

k〉.

Since the subgroup of a cyclic group is cyclic, H must be cyclic.

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(d) Suppose that Q is finitely generated. Then by part (c), Q must be cyclic. So,there exists some a

b∈ Q such that

Q = 〈ab〉

Choose x ∈ Q such that 0 < x < |ab|. Since a

bgenerates Q, there exists an integer

n > 0 such thatna

b= x,

but this means that |n||ab| = |na

b| = x < |a

b| =⇒ |n| < 1. This is a contradiction,

so Q is not finitely generated.

2.4.19 (a) Let ab∈ Q and k be a nonzero integer. Then a

kb∈ Q and

k( a

kb

)=

a

b

Thus, Q is divisible.

(b) Let G be a finite nontrivial group. Let k = |G| and let a ∈ G \ {1}. Then, for allx ∈ G,

xk = 1

so there is no x such that xk = a and G is not divisible.

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