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Subgroups and CosetsThroughout this section let G denote a group, Definition 1: We say that H is a subgroup of G if H is a group with respect to the multiplication · of G. Since · is already associative with respect to the elements of G, it is automatically associative for elements of H. Therefore, from the group axioms, we deduce that H is a subgroup of G if and only if H has the following properties SG1: H is closed under · ; that is, for SG2: SG3: If These three properties can be combined into a simple test for H to be a subgroup:
Proposition 2: Let Proof: There are two statements to prove. First assume that H is a subgroup of G and that Conversely, assume that Example 1: Let G denote the group of symmetries of the square. Two subgroups of G are given by
H_{1} = {I, R, R^{2}, R^{3}},
H_{2} = {I, F}.
Example 2: More generally, the dihedral group D_{n} contains two subgroups.
H_{1} = {I, R, R^{2},...,R^{n1}},
H_{2} = {I, F}.
Example 3: Let Then kZ is a subgroup of Z, considered as a group under addition. Example 4: The group C of continuous functions on R is a subgroup of the group of all functions on R. Example 5: Let f be the polynomial Theorem 3: If is any nonempty collection of subgroups of the group G, then the intersection
H^{*} =
is a subgroup of G. Proof: First observe that We shall now consider a general technique for manufacturing subgroups of a group G. Let S be any subset of G. Let be the collection of all subgroups containing S. is nonempty because
[S]^{ } =
is a subgroup of G and by the definition of , [S] contains S. Actually [S] is the smallest subgroup of G which contains S. For if H is a subgroup of G containing S, then
Definition 4: [S] is called the subgroup of G generated by S. If We can give another description of [S]. It is clear that [S] contains all products of the form (1)
a_{1}·a_{2}·a_{3}·...·a_{n},
where either
Definition 5: A group G is said to be cyclic if G is generated by a single element. That is, G is cyclic if there exists From our above characterization of [S] we see that G is cyclic if and only if there exists
G = {...,a^{2},a^{1},1_{G},a,a^{2},...}.
The element a is usually not unique. Moreover, the various powers of a need not all be distinct. We will study this possibility more closely below. Cyclic groups are the simplest groups. Later we will prove the fundamental theorem of abelian groups, which states that the cyclic groups are the building blocks from which abelian groups with a finite number of generators can be built. We have already met several examples of cyclic groups. Example 6: G = Z is a cyclic group. As a generator we may take 1. Example 7: G = Z_{n}(residue classes modulo n) is a finite cyclic group of order n. As a generator we may take 1. Example 8: Let G be an arbitrary group,
Definition 6: Let G be a group and let
Proposition 7: Suppose that a has finite order r. Then r is the smallest positive integer such that Proof: Since [a] is finite, not all of the powers a^{m}
n = pr + q (0 < q < r  1).
Then
= a^{n} = a^{pr+q} = (a^{r})^{p}a^{q}
= a^{q} (since a^{r} = 1).
= a^{j} = a^{r}a^{j} = a^{rj}
and
Proposition 8: Suppose that a has infinite order. Then
Proof: It suffices to consider the case
[a] = {1_{G} = a^{0}, a^{1},..., a^{r1}}.
(The powers of a may not all be distinct, however.) But this contradicts the fact that [a] is infinite. Moreover, if Combining Propositions 7 and 8 we can say something about cyclic groups. Let G be a cyclic group with generator a. There are two cases to consider Case 1: G is finite. Suppose that G has order r. Then by Proposition 7,
G = {1_{G}= a^{0}, a^{1}, a^{2},..., a^{r1}},
and a^{r} = 1_{G}. Case 2: G is infinite. In this case,
G = {...,a^{2}, a^{1}, a^{0} = 1_{G}, a^{1}, a^{2},...},
where all powers of a are distinct. Note that in case 1, G closely resembles Z_{r}, while in case 2, G closely resembles Z. In fact, in a later section, we will show that for purposes of group theory, any cyclic group is either Z of Z_{r}. Let us now return to the consideration of the subgroups of an arbitrary group G. One very important, but difficult, problem is to determine all of the subgroups of G. When G is finite, the problem can be solved by brute force. For we can list all of the subsets of G and then test each subset to determine whether it is a subgroup. But this solution to the problem is not very practical. Therefore, we must look for some simple criteria which can rule out most subsets of G from being subgroups. The simplest criterion is due to Lagrange and is one of the most fundamental theorems in the theory of groups. Theorem 9(Lagrange): Let G be a finite group, H a subgroup of G. Then the order of H divides the order of G. Before proving Lagrange's theorem, it will be necessary for us to build up some machinery. This machinery may seem very ad hoc at this time, but the ideas we will develop here will be quite natural and will be extremely important later when we define quotient groups Let
Ha = {ha  h H},
The set Ha is called a right coset of G with respect to H. We could just as easily consider left cosets of the form
aH = {ah  h H}.
But throughout this argument, we will stick to right cosets. We will denote the set of all right cosets of G with respect to H by H\G. The set of all left cosets will be denoted G/H. The element a is called a representative of aH. Lemma 10: All right cosets contain H elements. Proof: It is clear that Ha contains at most H elements. In order to prove that the number is exactly H, it suffices to show that all the elements
h_{1}·a = h_{2}·a (h_{1}, h_{2} H),
we have
Lemma 11: If Ha Hb , then Proof: Let g Ha Hb. Then
g = h_{1}· a = h_{2}· b (h_{1},h_{2} H)
Therefore,
Ha = Hh_{1}^{1}·h_{2}·b.
Since H is a subgroup,
Ha = Hb.
Every element of G is contained in some right coset. For example, a is contained in Ha. Let the number of distinct cosets of G with respect to H be k. Then if the right cosets are
Ha_{1}, Ha_{2}, ..., Ha_{k},
we can write (2)
G = Ha_{1} Ha_{2} ... Ha_{k}.
By Lemma 11, the union of (2) defines a partition of G and, by Lemma 10, each subset of the partition contains H elements. Proof of Lagrange's Theorem: Let
n = km HG.
Earlier we showed that with every partition of a set, there is associated an equivalence relation such that the subsets in the partition are precisely the equivalence classes of the equivalence relation. In the case of the partition (2), the equivalence relation is (3)
a ~ b b · a^{1} H.
Remark: Lagrange's theorem does not assert that if m is a positive integer which divides the order of G, then there exists a subgroup H of G having order m. In fact, this statement is false. Lagrange's theorem has a number of very important consequences. To mention but a few:
Corollary 12: Let G be a finite group, Proof: The order of a equals the order of the subgroup [a] of G. Therefore, the assertion follows directly from Lagarange's theorem.
Corollary 13: Let G be a finite group of order n, Proof: Let m denote the order of a. Then by Corollary 12,
a^{n} = a^{mk} = (a^{m})^{k} = 1_{G}^{k} = 1_{G}.
Corollary 14: A group of prime order is cyclic. Proof: Let G have a prime order p. Then the only positive divisors of p are 1 and p. Let
Corollary 15: Let n be a positive integer,
a^{(n)}1 (mod p).
where (n) denotes Euler's function. In particular, if p is prime and
a^{p1} 1 (mod p).
Proof: Let us apply Corollary 13 to the element a of the group Z_{n}^{x} of order (n). We get that
a^{(n)} = a^{(n)} = 1.
Therefore,
a^{(n)} 1 (mod n).
Corollary 15 was first discovered by Pierre Fermat in case n = a prime number and is known as "Fermat's little theorem." The general form of Corollary 15 is due to Euler. Let m be a positive integer. Since [a^{m}] is a subgroup of [a], Corollary 12 implies that the order of a^{m} divides the order of a. However, we can be much more specific:
Theorem 16: Let a have order n. then a^{m} has order n/(n,m). In particular if Proof: Let us first consider two special cases of the theorem: Case 1: mn. Suppose that Case 2:
msr + ntr = r
and
a^{r} = a^{msr} · a^{ntr}
=(a^{m})^{rs} · (a^{n})^{tr}
= 1_{G}.
Therefore, since a has order n, we see that General Case: Suppose that
m = m_{0}p, n = n_{0}p.
Then Definition 17: Let G be a group, and let H be a subgroup of G. The number of right cosets in H\G is called the index of H in G and is denoted [G:H]. The mapping Before leaving this section, it is important to reiterate the notion that the cosets of subgroup H of G form a partition of G. This is of fundamental importance in our studies. To summarize this fact we present the following theorem which is actually just rewording what was already stated. Proof of which is given by Lemma 10 and Lemma 11. Theorem 18: If H is a subgroup of G and aH and bH are cosets of H then (1) aH = bH or aH and bH are disjoint. (2) aH and bH are equivalent sets. 
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