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Subgroups of Groups
S. F. Ellermeyer
November 8, 2006
Suppose that G is a group and that H G. If H is also a group (under
the same operation as G), then we say that H is a subgroup of G. If G
is a group with identity element e, then E = feg is a subgroup of G called
the trivial subgroup of G. It is also true, for any group G, that G is a
subgroup of itself. If H is a subgroup of G with H 6= G, then H is called a
proper subgroup of G.
Example 1 The set of even integers is a proper subgroup of the group of all
integers under addition. The set of all multiples of 5 is also a proper subgroup
of the integers under addition. In fact, if k is any …xed integer, then
kZ = fkn j n 2 Zg = f: : : ; 3k; 2k; k; 0; k; 2k; 3k; : : :g
is a subgroup of the integers under addition. Note that if k = 0, then we
obtain the trivial subgroup f0g; whereas if k = 1, then we obtain the entire
group Z. Thus, unless k = 1, kZ is a proper subgroup of Z.
Example 2 Consider the multiplicative group Q f0g. A subgroup of this
group is H = f 1; 1g. Another subgroup of Q f0g is
H = f2n j n is an integerg =
1 1 1
: : : ; ; ; ; 1; 2; 4; 8; : : : .
8 4 2
To see this, note that H is closed under multiplication because if m and n
are any integers, then
2m 2n = 2m+n .
Also, H clearly satis…es the associative property (since Q does and H
1 2 H, and every member of H has a multiplicative inverse in H.
1
Q),
Example 3 Let us determine all of the subgroups of the group Z4 = f0; 1; 2; 3g
under addition. Of course, Z4 is a subgroup of Z4 and we also have the trivial
subgroup f0g. Suppose that 1 is in some subgroup, H, of Z4 . Then, since H
is closed (under addition), we see that 1 + 1 = 2 2 H, 1 + 2 = 3 2 H and also
0 2 H because the identity element must be in H (or because 1 + 3 = 0 2 H).
Thus, if 1 2 H, then it must be true that H = Z4 . Now suppose that H is a
subgroup of Z4 with 3 2 H. Then 3 + 3 = 2 2 H and 3 + 2 = 1 2 H and we
see once again that H = Z4 . The only non–trivial proper subgroup of Z4 is
H = f0; 2g.
Example 4 The group U8 = f1; 3; 5; 7g (under multiplication) has four non–
trivial proper subgroups. They are f1g, f1; 3g, f1; 5g, and f1; 7g. Notice that
f1; 3; 5g is not a subgroup of U8 because 3 5 = 7 so f1; 3; 5g is not closed
under multiplication. Likewise, f1; 3; 7g and f1; 5; 7g are not subgroups of
U8 .
The proof of the following Proposition is left as homework.
Proposition 5 If G is a group and H1 and H2 are subgroups of G, then
H1 \ H2 is a subgroup of G.
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Cyclic Groups
A group, G, is said to be cyclic if there exists an element a 2 G such that
G = fan j n is an integerg. (Note that if G is an additive group, then we
would write na instead of an .) If a is an element or G that has the property
described above, then we say that a is a generator of G and we write
G = hai.
Example 6 The group, Z, of integers under addition is a cyclic group with
generator 1. This is because, for any integer n, we have n = n 1. Note that
1 is also a generator of Z. Thus we could write Z = h1i or Z = h 1i. Z
does not have any generators other than 1 and 1.
Example 7 The group, 2Z, of even integers is cyclic with generator 2 because any even integer can be written in the form n 2 where n is an integer.
Thus 2Z = h2i. (It is also true that 2Z = h 2i.)
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Example 8 The group, Z4 = f0; 1; 2; 3g, under addition is cyclic with generator 1 because
1=1
1+1=2
1+1+1=3
1 + 1 + 1 + 1 = 0.
Thus every element of Z4 can be written as n 1 for some integer n. Does Z4
have any other generators?
2=2
2+2=0
2+2+2=2
etc.
shows that 2 is not a generator of Z4 .
3=3
3+3=2
3+3+3=1
3+3+3+3=0
shows that 3 is a generator of Z4 . Thus Z4 = h1i = h3i.
Exercise 9 It is clear that, for any integer n 1, the additive group Zn is
cyclic with generator 1. Find all of the generators of Z5 , Z6 , Z7 , and Z8 .
Can you make a general conjecture about the generators of Zn ?
Example 10 The group U8 = f1; 3; 5; 7g under multiplication is not cyclic
because 32 = 1, 52 = 1, and 72 = 1 so neither 3, nor 5, nor 7 generates U8 .
Even though not all groups are cyclic, all groups contain at least one
cyclic subgroup –the subgroup E = feg = hei. In general, if G is any group
and a is an element of G, then hai is an abelian subgroup of G (even if G
itself is not abelian). The subgroup hai is called the cyclic subgroup of G
generated by a. Of course, it may be the case that hai = G (in which case G
itself is cyclic) or it may be the case that hai = hbi for two di¤erent elements,
a and b, of G.
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Proposition 11 Let G be a group and let a 2 G. Then hai = fan j n is an integerg
is an abelian subgroup of G.
Proof. Since, for any integers m and n, we have am an = am+n , we see that
hai is closed under the operation of G. Also, since G obeys the associative
law, then so does hai just by virtue of the fact that hai
G. In addition
a0 = e 2 hai, and for any member, an , of hai, we see that a n 2 hai and a n
is the inverse of an . This proves that hai is a group (and hence a subgroup
of G.) To see that hai is an abelian group, note that for any two integers m
and n, we have am an = an am .
Example 12 Consider the dihedral group D3 whose Cayley table is shown
below. Find the cyclic subgroups of D3 that are generated by each of the
members of D3 .
*
R0
R1
R2
Fa
Fb
Fc
R0
R0
R1
R2
Fa
Fb
Fc
R1
R1
R2
R0
Fb
Fc
Fa
R2
R2
R0
R1
Fc
Fa
Fb
Fa
Fa
Fc
Fb
R0
R2
R1
Fb
Fb
Fa
Fc
R1
R0
R2
Fc
Fc
Fb
Fa .
R2
R1
R0
Solution: The subgroup generated by R0 is hR0 i = R0 . For R1 we have
R11 = R1
R12 = R2
R13 = R0
so
hR1 i = fR0 ; R1 ; R2 g .
Likewise we see that
hR2 i = fR0 ; R1 ; R2 g .
Now note that
Fa1 = Fa
Fa2 = R0
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so
hFa i = fR0 ; Fa g .
Likewise it can be seen that hFb i = fR0 ; Fb g and that hFc i = fR0 ; Fc g. Thus
D3 has one cyclic subgroup of order three and three cyclic subgroups of order
two. (The cyclic subgroup of order three is generated by two di¤erent elements
of D3 .)
Example 13 Consider the multiplicative group Q
subgroup of Q generated by 1=2.
Solution: Observe that
1
2
0
1
2
1
1
2
2
f0g. Find the cyclic
=1
=
1
2
1
4
etc.
=
and
1
2
1
1
2
2
=2
=4
etc.
Thus
1
2
=
1 1 1
: : : ; ; ; ; 1; 2; 4; 8; : : : .
8 4 2
Proposition 14 Suppose that G is a group of …nite order with identity element e and suppose that a 2 G. Then there exists an integer k > 0 such that
ak = e.
Proof. If a = e, then a1 = e. Thus suppose that a 6= e. Since G is of …nite
order, it is not possible that all of the powers a1 ; a2 ; a3 ; . . . are di¤erent
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from one another (for then G would have in…nite order). Therefore, there
must exist some positive integers m and n with m > n such that am = an .
However this implies that am n = a0 = e (and note that m n is a positive
integer). This proves the assertion.
Corollary 15 If G is a group of …nite order and H is a non–empty subset
of G that is closed under the operation of G, then H is a subgroup of G.
Proof. It is clear that H satis…es the associative property (because G does
and H
G). Also, H 6= ; so there exists an element a 2 H. By the
preceding Proposition (since G is of …nite order), there must exist some
positive integer k such that ak = e. Since H is closed under the operation of
G, it must therefore be true that e 2 H. Finally, we will show that if a 2 H,
then a 1 2 H. If a = e, then a 1 = e so a 1 2 H. If a 6= e, the let k be a
positive integer such that ak = e. Since a 6= e, we know that k
2. Also
a 1 ak = a 1 e and thus a 1 = ak 1 . Since H is closed under the operation
of G, it must be the case that ak 1 2 H. Therefore a 1 2 H. This proves
that H is a subgroup of G.
De…nition 16 Suppose that G is a group and suppose that a 2 G. We de…ne
the order of a, denoted by jaj, to be the order of the cyclic subgroup generated
by a. Thus jaj = jhaij. (Note that jaj is either a positive integer or in…nity.)
Example 17 In Example 12, we learned that jR1 j = jR2 j = 3, whereas
jFa j = jFb j = jFc j = 2 in D3 . In Example 13, we learned that j1=2j = 1 in
Q f0g.
The following corollary to Proposition 14 insures that if G is a group of
…nite order and a 2 G, then the entire subgroup hai is actually generated by
positive powers of a.
Corollary 18 If G is a group of …nite order and a 2 G, then
hai = fan j n is a positive integerg :
Proof. We know (by de…nition) that
hai = fan j n is an integerg .
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Let
A = fan j n is a positive integerg .
We want to prove (under the assumption that G is of …nite order) that
hai = A. First we will dispense with the case that a = e. In this case, it is
clear that hai = feg and also that A = feg. Therefore hai = A. Henceforth
we assume that a 6= e.
It is clear that A hai. Thus we only need to prove that hai A.
Let x 2 hai. Then there exists an integer m such that x = am . If m > 0,
then x 2 A. Thus suppose that m
0. By Proposition 14, we know that
there exists a positive integer k such that ak = e. Since k is a positive integer,
we can …nd a positive integer r such that rk > m. (Note that m
0
r
because we are assuming that m 0.) Since ark = ak = er = e, we obtain
x = am = eam = ark am = ark+m
and since rk + m > 0 we see that x 2 A. Therefore hai
now proved that hai = A.
2
A and we have
Product Groups
Suppose that G1 is a group with operation 1 and that G2 is a group with
operation 2 . Then we can from a new group called the product group of
G1 and G2 by using the Cartesian product G1 G2 as the underlying set and
de…ning the operation, , on G1 G2 as follows:
(a; b) (c; d) = (a
1
c; b
2
d) .
The group G1 G2 is also called the direct product of G1 and G2 . If
additive notation is in order (due to the fact that additive notation is being
used for both G1 and G2 ), then we usually write G1 G2 instead of G1 G2
and we call G1 G2 the direct sum of G1 and G2 .
Example 19 Let us consider the direct sum Z2 Z3 . Since Z2 = f0; 1g and
Z3 = f0; 1; 2g, the elements of Z2 Z3 are (0; 0) ; (0; 1) ; (0; 2) ; (1; 0) ; (1; 1) ;
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and (1; 2). The Cayley table for Z2
*
(0; 0)
(0; 1)
(0; 2)
(1; 0)
(1; 1)
(1; 2)
(0; 0)
(0; 0)
(0; 1)
(0; 2)
(1; 0)
(1; 1)
(1; 2)
(0; 1)
(0; 1)
(0; 2)
(0; 0)
(1; 1)
(1; 2)
(1; 0)
Z3 is as follows:
(0; 2)
(0; 2)
(0; 0)
(0; 1)
(1; 2)
(1; 0)
(1; 1)
(1; 0)
(1; 0)
(1; 1)
(1; 2)
(0; 0)
(0; 1)
(0; 2)
(1; 1)
(1; 1)
(1; 2)
(1; 0)
(0; 1)
(0; 2)
(0; 0)
(1; 2)
(1; 2)
(1; 0)
(1; 1) .
(0; 2)
(0; 0)
(0; 1)
Let us compare the above group with the dihedral group D3 which is also a
group of order 6 (and whose table is given in Example 12). D3 has one cyclic
subgroup of order three and three cyclic subgroups of order two. In Z2 Z3 ,
we have
2 (0; 1) = (0; 2)
3 (0; 1) = (0; 0)
so j(0; 1)j = 3. Likewise j(0; 2)j = 3 and we see that f(0; 0) ; (0; 1) ; (0; 2)g is
a subgroup of order 3 in Z2 Z3 .
Also
2 (1; 0) = (0; 0)
so j(1; 0)j = 2 and f(0; 0) ; (1; 0)g is a subgroup of order 2 in Z2 Z3 . However,
Z2 Z3 does not have any more cyclic subgroups of order 2 because
2 (1; 1) = (0; 2)
3 (1; 1) = (1; 0)
4 (1; 1) = (0; 1)
5 (1; 1) = (1; 2)
6 (1; 1) = (0; 0)
and this shows that j(1; 1)j = 6 and hence that Z2 Z3 = h(1; 1)i (meaning
that Z2 Z3 is in fact a cyclic group). It can also be con…rmed that j(1; 2)j =
6. We conclude that D3 and Z2 Z3 (although both groups of order 6) are
essentially di¤erent from each other. Z2 Z3 is cyclic and D3 is not. Also,
Z2 Z3 is abelian and D3 is not.
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We note, referring to the above example, that Z2 Z3 has subgroups
that are “copies” of Z2 and Z3 . In particular f(0; 0) ; (1; 0)g is a copy of Z2
and f(0; 0) ; (0; 1) ; (0; 2)g is a copy of Z3 . This is always the case with direct
products (or direct sums). In G1 G2 (with respective identity elements e1
and e2 ), the subgroup f(x; e2 ) j x 2 G1 g is a copy of G1 and the subgroup
f(e1 ; y) j y 2 G2 g is a copy of G2 . More generally, in a direct product G1
G2
Gn , the subgroup f(e1 ; e2 ; : : : ; x; : : : ; en ) j x 2 Gi g is a copy of
Gi . Because of this fact, for any two positive integers m and n such that
n is divisible by m, we can always construct a group of order n that has a
subgroup of order m.
Example 20 Suppose that we would like to construct a group of order 24
that contains a copy of the Klein 4–Group, K. Since jKj = 4 and 24 is
divisible by 4, then this construction is possible. We can choose any group
of order 6, say D3 , and be assured that D3 K is a group of order 24 that
contains a copy of K. (Of course, D3 K also contains a copy of D3 .)
We conclude with a theorem that gives the order of an element in a
product group in terms of the orders of the components of this element in
their respective groups.
Theorem 21 Let G1 ; G2 ; : : : ; Gn be groups of …nite order and let ai 2 Gi
for each i = 1; 2; : : : ; n. Then the order of the element (a1 ; a2 ; : : : ; an ) in the
product group G1 G2
Gn is the least common multiple of the orders
of a1 ; a2 ; : : : ; an in the respective groups Gi ; i = 1; 2; : : : ; n.
In order to prove Theorem 21, we will need a Lemma that is of interest
in its own right.
Lemma 22 Suppose that G is group and suppose that a 2 G with jaj = n.
Then am = e if and only if m is divisible by n.
Proof. Suppose that m is divisible by n. Then m = qn where q is an integer.
This implies that
am = aqn = (an )q = eq = e.
Now suppose that m is not divisible by n. Then m 6= 0 and m 6= n. If
0 < m < n, then am 6= e because this would contradict the fact that jaj = n.
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If m > n, then the Division Algorithm gives us integers q and r such that
m = qn + r and 0 < r < n. In this case we have
am = aqn+r = (an )q ar = eq ar = ear = ar
and we know that ar 6= e (because 0 < r < n) and thus am 6= e. Finally,
if m < 0, then m > 0 and we thus know that a m 6= e. But this implies
that am 6= e (because if am = e then we must have a m am = a m e and hence
e = a m ).
We now give the proof of Theorem 21. We will prove the theorem only
for the case of two groups, G1 and G2 .
Proof of Theorem 21: Suppose that G1 and G2 are groups of …nite
order (with respective operations 1 and 2 and respective identity elements
e1 and e2 ) and suppose that a1 2 G1 and a2 2 G2 . Suppose also that ja1 j = s
and ja2 j = t. We want to prove that j(a1 ; a2 )j = lcm (s; t).
To begin, note that as1 = e1 and at2 = e2 . Letting m = lcm (s; t), we
obtain
m
(a1 ; a2 )m = (am
1 ; a2 ) = (e1 ; e2 )
(by the preceding lemma, because m is divisible by both s and t). To prove
that j(a1 ; a2 )j = m, we must show that (a1 ; a2 )n 6= (e1 ; e2 ) for any integer n
such that 0 < n < m. To this end, suppose that 0 < n < m and suppose
that (a1 ; a2 )n = (e1 ; e2 ). Then (an1 ; an2 ) = (e1 ; e2 ) which means that an1 = e1
and an2 = e2 . By the preceding lemma, it must then be the case that n is
divisible by both s and t or, in other words, that n is a common multiple of
s and t. Since the least common multiple of any two numbers divides any
common multiple of those two numbers, it must then be the case that n is
divisible by m. However this is a contradiction because 0 < n < m. We thus
conclude that j(a1 ; a2 )j = m.
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