DIRECT PRODUCTS OF GROUS

In this chapter, we pay special attention to finite groups.

Definition 3 – 1 .
Let (H,*) and (K,*) be normal subgroups of the group (G,*). . Then G is said to be the INTERNAL DIRECT PRODUCT . of H and K , written
G =H ? K, if G = H * K and H ? K = {e}.

Before attempting to set out the theory of direct products . . in an orderly fashion let us look at two examples.

Example 1 .
Consider the Klein 4 – group K4 = {e, a, b, ab}, where . a2 = b2 = e and ab = ba. . Since K4 is commutative, all its subgroups are normal.
If H = (a) and K = (b) ,
then conditions of Definition 6 - 1 are satisfied.
Thus K4 = H ? K is the internal direct product of two cyclic . subgroups of order 2.

Note
Some groups cannot be expressed as the internal direct . . product of two nontrivial normal subgroups (such groups . . are referred to as INDECOMPOSABLE ).

Example 2 .
Let G be any infinite cyclic group G = (a) .
Suppose that G = H ? K , where H , K are nontrivial subgroups . of G. Then H and K are cyclic .
If H = (an) and K = (am) , then anm ? H? K .
Hence H? K ? {e}, which is contradiction .
If anm = e , then G is finite .

Exercise .
The group (Z15 , +15) is internal direct product of { 0,5,10} and {0,3,6,9,12}.
b) The group of symmetries of the square is . indecomposable. 27

There exist several criteria for a group to be an internal . direct product of its subgroups. We establish one such
. condition below.

Theorem 3 – 1 .
The group (G , *) is the internal direct product of its . subgroups H and K if and only if
Each element x ? G can be uniquely expressed in the form x = h * k , where h? H and k ? K and
Any element of H commutes with any element of K ( h*k = k * h).
Proof .
Suppose that G = H ?K , then G = H * K,and any x? G
x = h * k , with h ?H and k ? K .
To see the representation is uniquely ,
Let x = h1 *k1 for h1 ? H and k1 ? K .
Then h * k = h1 * k1,
Hence h1-1 *h = k1* k-1.
Since h1-1 *h ? H and k1* k-1? K ,
Then each of them is in H and K .
Since H? K = {e} , then h1-1 *h = e = k1* k-1
So h = h1 and k = k1 . Hence (1) is proving .
Let h ?H and k ? K and consider [h , k ] = hkh-1k-1 .
Since K is normal , then hkh-1? K , hence (hkh-1)k-1 ? K.
Similarly h(kh-1k-1)? H .
So hkh-1k-1 ? H? K
Then hkh-1k-1 = e, so hk = kh . Hence (2) is proving
For the other direction of the theorem , to prove that . G = H? K When the conditions (1) and (2) are hold .
From condition (1) G = H * K
Pick any h?H and x ? G . Then x=h1k1 for h1?H , k1? K
xhx-1 =(h1k1)h(h1k1)-1 = h1 h h1-1? H ?
So H ? G. Similarly K? G.
To complet the proof , we must show that H? K = {e} .
If x ? H? K , then
x = xe (x? H , e? K) and x = ex ( e?H , x? K)
From (1) x= e .
Hence H? K = {e} .

28
DIRECT PRODUCT AND QUOTIENT GROUP

Now we will study the relation between the internal direct
product and quotient group .

Theorem 3 - 2 .
If H and K are normal subgroups of the group G such that
G = H ? K , then G/H ? K and G / K ? H .
Proof .
Conceder f : G ? K defined as :
If x = hk , then f (x) = k .
The uniqueness of the representation for x assure that f is well
defined .
f is onto?
If x = hk and y = h1k1,
Then xy = (hk)(h1k1) = (hh1)(kk1) ?
Hence f(xy) = kk1 = f(x) f(y), then f is homomorphism .
ker(f) = {hk? HK | f(hk) = e }
= {hk? HK | k = e } = H .
So G / H ? K ?
The proof of G / K ? H is home work .
Note .
The Second Isomorphism Theorem is via with Theorem 3 – 3
G / H = HK / H ? K / H? K = K / {e} = K .
Also G / K ? H .

Corollary .
If G is finite group and if G = H ? K , then
o(G) = o(H) o(K) .
Proof .
From Lagrange , s Theorem
o(G ) = o(H) [G/H]
But [G : H ] = o(K) ?
Hence o(G) = o(H) o(K) .
Example .
H = {0 , 5 } and K = { 0 , 2 , 4 , 6 , 8 } are normal subgroups of
of (Z10 , +10)?
Z10 = H? K ?
Z10 / H ? K and Z10 / K ? H ?
o(Z10) = 10 = 2 . 5 = o(H) o(K) .
Also G / K ? H .
29

Lemma .
If H and K are finite subgroups of the group G , then

o(H) o(K)
o(HK) = ـــــــــــــــــــ
o( H? K )
Proof .
From definition of HK o(HK)? o(H) o(K) .
Let hk ? HK .
If hk = h1k1 h1? H , k1? K , where h1? h or k1? k .
Suppose that x? H? K and hk ? HK .
hk = (hx)(x-1k) ? HK ?
So hk is duplicated in HK at least o(H ? K) times .
If hk = h1k1 , then x = h-1h1 = kk1-1 lies in H ? K .
Hence h1 = hx , k1 = x-1k .
Then hk appears in HK exactly o(H?K) times.
Hence the number of distinct elements in HK is o(H)o(K)
divided by the number of times each element appears [o(H? K)].