Course 111: Algebra, 17th November 2006
1. Suppose G is a cyclic group of order 4, G = {e, a, a2 , a3 } and a4 = e.
Find the order of each element of G. Find all the subgroups of G.
List the distinct left cosets of the subgroup H = {e, a2 } in G.
Solution
o(e) = 1 since e1 = e
o(a) = 4 since a.a.a.a = a4 = e
o(a2 ) = 2 since a2 .a2 = a4 = e
o(a3 ) = 4 since a3 .a3 .a3 .a3 = (a4 )3 = e
The only proper subgroup of G is {e, a2 }. ie it contains the group identity element, e, it is closed under the group operation, a2 .e = a2 , a2 .a2 =
a4 = e, e.e = e and each element has an inverse since e.e = e so e−1 = e
and a2 .a2 = e so (a2 )−1 = a2 .
In addition, {e}, {e, a, a2 , a3 } are also (improper) subgroups.
Given H = {e, a2 } the left cosets are {e, a2 }, {a, a3 }, {a2 , a4 = e}, {a3 , a5 =
a} so the two distinct left cosets are
{e, a2 } and {a, a3 }.
2. Prove that for H and K both subgroups of a group G. The product,
HK is itself a subgroup of G iff HK = KH.
Note: HK = {x ∈ G|x = hk, h ∈ H, k ∈ K}
Proof:
Begin by supposing that HK = KH - then we must prove that HK is
a subgroup.
HK = KH means that if h ∈ H and k ∈ K then hk = k1 h1 for some
k1 ∈ K and h1 ∈ H. Note this doesn’t require that k1 = k or that
h1 = h. To prove that HK a subgroup we have to show it is closed
and that every element in HK has an inverse in HK.
• closure: suppose x = hk ∈ HK and y = h0 k 0 ∈ HK. Then xy =
hkh0 k 0 . But now since kh0 ∈ KH = HK we have kh0 = h2 k2 with
h2 ∈ H and k2 ∈ K. Hence xy = h(h2 k2 )k 0 = (hh2 )(k2 k 0 ) ∈ HK
and so HK is closed.
• inverses: x−1 = (hk)−1 = k −1 h−1 ∈ KH = HK. So, x−1 ∈ HK.
Then, HK a subgroup.
Now suppose HK is a subgroup - then we must show HK = KH.
If HK a subgroup of G then for any h ∈ H and k ∈ K, h−1 k −1 ∈ HK
and so kh = (h−1 k −1 )−1 ∈ HK. Thus KH ⊂ HK.
Now if x is any element of HK then x−1 = hk ∈ HK and so x =
(x−1 )−1 = (hk)−1 = k −1 h−1 ∈ KH and so HK ⊂ KH.
Thus, HK = KH.
3. Recall the properties of cosets listed in your notes. From this list prove
the following:
For a subgroup H of a group G,
• x ∈ xH for all x ∈ G.
Proof: Consider x ∈ G. Then x = xe for the identity element, e
of G. But e ∈ H. Therefore x ∈ xH.
• if x and y ∈ G and if y = xa for some a ∈ H then xH = yH.
Proof: Let x, y ∈ G where y = xa and a ∈ H. Then yh = x(ah)
and xh = y(a−1 h) for all h ∈ H. Also, ah ∈ H and a−1 h ∈ H
for all h ∈ H since H a subgroup of G. Therefore yH ⊂ xH and
xH ⊂ yH and hence xH = yH.
• each left coset of H in G has the same number of elements as H.
Proof: Suppose that xH ∩ yH is non-empty for x, y ∈ G. And
let z ∈ (xH ∩ yH). Then z = xa for some a ∈ H and z = yb for
some b ∈ H. It follows from the second proof that zH = xH and
zH = yH. Therefore xH = yH.