Page 47 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes Now, 0  mZ. Therefore, mZ  . Also, for mr, ms  mZ, mr-ms = m(r-s)  mZ.
Therefore, mZ is a subgroup of Z.

Note that m is the least positive integer in mZ.
Now, let H  (0) be a subgroup of Z and S = { i | i > 0, i  H).
Since H # {0), there is a non-zero integer k in H. If k > 0, then k  S. If k < 0, then (-k)  S, since
(–k)  H and (–k) > 0.
Hence, S   .
Clearly, S  N. Thus, by the well-ordering principle S has a least element, say s. That is, s is the
least positive integer that belongs to H.
Now sZ  H. Why? Well, consider any element st  sZ.
If t = 0, then st = 0  H.
If t > 0, then st = s + s + ..... + s (t times)  H.

If t < 0, then st = (–s) + (–s) + ....+ (–s) (–t times)  H.
Therefore, st  H  t  Z. That is, sZ  H.
Now, let m  H. By the division algorithm m = ns + r for some n, r  Z, 0  r < s. Thus, r = m – ns.
But H is a subgroup of Z and m, ns  H. Thus, r  H. By minimality of s in S, we must have r = 0,
i.e., m = ns. Thus, H  sZ.

So we have proved that H = sZ.
You know that the polar form of a non-zero complex number z  C is z = r (cos + i sine), where
r = | z | and  is an argument of z. Moreover, if  , is an argument of z and  that of z . then  l
1
2
1
2
+  is an argument of z z . Using this we will try to find the nth roots of 1, where n  N.
2
l
2
If z = r (cos  + i sin !) is an nth root of 1, then z = 1.
n
Thus, by De Moivre’s theorem,
1 = z = r (cos n + i sin n), that is,
n
n
cos (0) + i sin (0) = r (cos n + i sin n). ................. (1)
n
Equating the modulus of both the sides of (1). we get r = 1, i.e., r =l.
n
On comparing the arguments of both sides of (1), we see that 0 + 2nk (k  Z) and n are
arguments of the same complex number. Thus, n can take any one of the values 2k, k  Z. Does
2nk
this mean that as k ranges over Z and  ranges over we get distinct nth roots of 1? Let us
n
2nk 2zk 2 m 2 m 2nk 2nm


find out. Now, cos + i sin = cos  isin if and only if   2nt for
n n n n n n
some t  Z. This will happen iff k = m + nt, i.e., k  m (mod n). Thus, corresponding to every r
2 r 2 r


r
in Z, we get an nth root of unity, z = cos  isin , 0   n; and these are all the nth roots
n n
of unity.
For example, If n = 6, we get the 6th roots of 1 as z , z , z , z , z and z , where zj =
2
l
3
5
4
0
2 j 2 j


cos  isin , j  0,1,2,3,4,5. In Figure 3.1 you can see that all these lie on the unit circle (i.e.,
6 6
the circle of radius one with centre (0, 0)). They form the vertices of a regular hexagon.
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