Suppose that a group contains elements of orders 1 through 10.
Answers
he following are from Gallian, Chapters 7 and 9. The numbering is first given for the 7th edition, and
is then followed by the 6th edition’s numbering (if the numbering is different).
• # 7.15 : Suppose | G |= pq with p and q prime numbers. By Lagrange’s Theorem, the possible
subgroups of G have orders 1, p, q, or pq. For a subgroup H of G to be proper, it must have order
less than G, so we can immediately eliminate pq. In the case that | H |= 1, H contains only e so it
is clearly cyclic. In the case that | H |= p or | H |= q, then H has prime order, so it must be cyclic.
This completes the argument.
• # 7.20 : Since H and K are both subgroups of G, we know that H ∩ K must contain e, so it is
nonempty. Suppose there is some other element, g, that is in H ∩ K. Then g ∈ H and g ∈ K, so
| g |> 1 must divide the order of H and of K, since both H and K are themselves groups. But, since
| H |= 12 and | K |= 35 have no common divisors greater than 1, we have derived a contradiction.
Thus, the only element in H ∩ K is e, so | H ∩ K |= 1.
• # 7.39 (resp. # 7.35 ): Since G has elements of orders 1 through 10, it’s clear that if | G |= k, then
10 | k, 9 | k, ..., 2 | k. Thus, the least possible order of G is lcm(1, 2, 3, 4, 5, 6, 7, 8, 9, 10) which is
2
3
· 3
2
· 5 · 7 = 2520.
• # 9.1 : No, H is not normal in S3. Consider that (13)H = {(13),(123)} 6= H(13) = {(13),(132)}.
• # 9.7 : Suppose that H has index 2 in G. Then, for any a not in H, we have that aH ∩ H = ∅ and
aH ∪ H = G. Therefore, the complement of H in G is simply aH. The exact same argument where
Ha is considered instead of aH shows that the complement of H also equals Ha. Thus aH = Ha for
all a not in H. For any b ∈ H, bH = H = Hb. Thus, for all g ∈ G, we have gH = Hg, so that H is
normal.
• # 9.8a: Referencing Table 5.1, consider that (123)H = {(123),(134)} 6= H(123) = {(123),(243)}.
• # 9.11 : No. Consider D3 and the subgroup H that contains only the rotations. Note that H has
index | D3 | / | H |= 6/3 = 2, and is therefore normal by 9.7. The subgroup of rotations is clearly
Abelian. Furthermore, D3/H contains just two elements, and every two element group is cyclic and
therefore Abelian. So, both H and D3/H are Abelian, but D3 is non-Abelian.
• # 9.12 : Let G be an Abelian group and let H be a normal subgroup of G. Let x and y be in
G/H. Then x = aH = Ha and y = bH = Hb for some pair of elements a, b ∈ G. Therefore,
xy = (aH)(bH) = (ab)H = (ba)H = (bH)(aH) = yx, and we conclude that G/H is Abelian.
• # 9.37 : If we let the order of g ∈ G be n, then (gH)
n = g
nH = eH = H. Since H = eH is the
identity in G/H, we know that the order of gH must divide n. But this says exactly that the order of
gH in G/H divides the order of g in G.
• # 9.52 (resp. # 9.48 ):
– Part a:
· 1 i j k -1 -i -j -k
1 1 i j k -1 -i -j -k
i i -1 k -j -i 1 -k j
j j -k -1 i -j k 1 -i
k k j -i -1 -k -j i 1
-1 -1 -1 -j -k 1 i j k
-i -i 1 -k j i -1 k -j
-j -j k 1 -i j -k -1 i
-k -k -j i 1 k j -i -1