We will continue our study of Structure of
finites of a groups of a multiplicative group
of a field and today we will prove that these
subgroups are cyclic.
!
So (what) we will prove that if K is any field
and G is a subgroup of K cross, K cross is
a multiplicative group of the field K and
if G is finite then we want to prove 
G is cyclic, that means we want to prove that
G is generated by one element that is in our
notation G is H of y for some y, such a y
may not be unique and such a y is also called
a generator of G or also called primitive
element 
of G. So cyclic group has primitive element
and there may not be unique, there may be
many primitive elements this what we want
to prove but for the proof of this we will
need some preliminary results on finite groups
that I will recall, some of the my will recall
with the sketch of proofs and some of the
my will recall without proofs.
So for example let us started typical finite
group for example we have seen Z modulo m,
this is modulo m operations, congruence modulo
m. So there two operations plus and dot And
with respective that we have seen it is a
ring and whenever we have a ring we talk about
the unit group of the ring that is usually
denoted in this case Z m cross this means
you take the multiplicative monoid Z m dot
And take units in multiples element in their
monoid that is obviously a unit that obviously
subgroup of the multiplicative monoid of Z
mode m this is called group of units, group
of units of Z mode m and for example when
m is 0 hiss Z mode m is Z and therefore Z
mode m cross is Z cross which is only plus
minus 1.
So it is obviously cyclic of order two more
generally if m is positive remember also when
we did the definition etcetera for Z Mode
m, Z mode m as also same as Z additive group
of integers a modulo the subgroup generated
by m that is this, so this are the cosets
and there given by the remainders of when
you divide by m so there precisely m remainders
from 0 to m minus 1, so therefore elements
of Z mode m, Z m also we are denoting this
as residues that or equivalence classes so
the notation is this, these are precisely
the residues and there are precisely residues
are from 0 to m minus 1, so these are precisely
equivalence classes of 0, 1, 2 up to m minus
1 with the congruence modulo m relation, so
when m is positive the units are precisely
Z m cross these are precisely all those remainders
and we can assume them they are from 0 to
m minus 1 and gcd of a and m should be 1 they
are coprime to m, they are precisely the units
in Z mode m and then how many of them are
there that this is a very famous notation
for that due to Euler.
That is say that cardinality of Z m cross
is precisely phi of m where this phi is Euler’s
totient function this is Euler’s totient
function which maps any natural number m to
phi m and phi m is by definition cardinality
of all the integers, all the natural numbers
a, a 0 less equal to a less than m and gcd
of a and m the number of coprime integers
that is the phi of m, so the first thing to
note is that because this is a group and order
is phi m dividing any element there any element
will look like any element is a residue and
then if I raise to the power phi power m that
should be 1 in these in Z mode m but this
is precisely writing this is equivalent to
writing that if I raise a to the power phi
m that is congruent to 1 mode m this is also
called Euler’s theorem, so this is under
the assumption that gcd of a and m is 1, so
this is Euler’s theorem.
(pa) Particular case we have seen earlier
that is the famous Fermat’s little theorem
that says m is p here now any a smaller than
p and bigger equal to 1 these are coprime
to p does not divide a so therefore a power
p minus 1 is congruent to 1 mode p note that
in this case phi p is p minus 1 because everybody
is coprime to p, smaller than p so that is
the Fermat’s little theorem, ok now.
Now I want to recall what is called Chinese
remainder theorem so Chinese remainder theorem
this is about the solutions of simultaneous
congruences simultaneous solution of a simultaneous
congruences relations, so let m 1 to m r be
positive integers N plus which are pairwise
relatively prime 
that simply means if I take any pair m i and
m j they do not have any common factors that
is gcd is 1 for all i not equal to j and let
us take m is the product m 1 to m r, then
for every tuple of integers r tuple a 1 to
a r of integers the simultaneous congruences
a 1 X equal to X congruent to a 1 mode m 1,
X congruent to a r mode m r have a solutions
in Z. this simply means there exist in integers
X so that X is congruent to a 1 mode m 1,
X is congruent to a 2 mode m 2 and so on X
is congruent to a r mode m r, definitely there
is a solution moreover a solution is uniquely
determined modulo m solution may not be unique
but when you go mode m that is unique solution.
So this in the notation we can simply reformulate
neatly as follows.
So reformulation of CRT Chinese remainder
theorem the canonical ring homomorphism chi
bar which is homomorphism from Z mode m to
Z mode m 1 cross cross cross Z mode m r, this
is a product ring the component wise operations
so any residue here a suffix m this maps to
take that a and take the residues mode m 1
etcetera mode m r this is a canonical map
you just taking the residues this is an a
isomorphism of a ring and surjetivity is what
the existence of solution and the injectivity
of this map is precisely uniqueness model,
so this is also I will keep calling as Chinese
remainder theorem, so proof is not very difficult
so proof I am not going to prove this but
I will note down some corollaries.
So for example, 1 corollary the canonical
these canonical ring homomorphism chi bar
induces an isomorphism of unit groups 
that means this chi bar was from Z mode m,
so the unit group of so that notation is chi
bar x this is Z mode m x to obviously the
product ring the units are precisely the units
in each component that means these are the
this is a unit group of that product ring
this is an isomorphism because at the ring
level it is isomorphism so unit level it is
an isomorphism.
So in particular 
the order of this group that we know it is
phi m which is the order of the group Z mode
m cross but that is same as the order of this
group Z mode m 1 cross product Z mode m r
cross but these products is obviously the
product of the unit group of Z mode m 1 etcetera
Z mode m r but obviously these is phi m 1
etcetera phi m r so therefore this means that
phi is multiplicative function that means
whenever you have relatively prime integers
and when you apply phi to their product it
is the product of the phis.
So that was the (consequence) that is the
property of the Euler’s phi function.
So further if I specialized if I take precisely
if I take m equal to look at the prime decomposition
of m p 1 alpha 1, p r alpha r where p 1 to
p r are distinct 
primes and alpha 1 to alpha r are positive
natural numbers then Z mode m cross is precisely
Z mode p 1 to the alpha 1 cross cross Z mode
p r to the alpha r cross therefore their orders
are equal that is phi m equal to modular of
these group that is obviously p 1 to the alpha
1 minus 1 times p 1 minus 1 and so on p r
to the alpha r minus 1 p r minus 1 so nice
formula for phi, also we can also formulate
the Chinese remainder theorem completely in
group theoretic terms so let me do it, so
group theoretic formulations of CRT Chinese
remainder theorem, so that is a following
suppose you have G 1 to G r are finite groups
and let us denote G to with a product group
with the component wise binary operation product
group, ok then G is cyclic product group is
cyclic that means it has primitive element
if and only if a each factor every factor
G 1 to G r are cyclic and there orders are
coprime and G 1 to order of G 1 to order of
G r are pairwise relatively prime.
So that this again I will not prove but this
depends on this one can prove this I will
just make one comment this can proved easily
(by using) by induction proof by induction
by induction on r 
and the observation following observation
about the orders if g and h are two elements
in a group commuting elements of positive
orders 
then order of the product equal to product
of the orders if and only if order g and order
h are relatively prime this is this observation
is very simple very easy to prove, so I leave
it for you to check and now with this I will
prove the theorem I wanted to prove.
So that is the theorem every finite subgroup
G of K cross where K is a field is cyclic,
these what we wanted to prove, ok. K may not
be a finite field, K may be infinite field
also for example K equal to C or K equal to
R, so proof 
so let us look at the order of G, m is the
order of G which is I will write other prime
decomposition p 1 alpha 1, p r alpha r where
these p 1 to p r are distinct primes 
and alpha 1 to alpha r positive natural numbers.
So I will call these p 1 to the alpha 1 as
m 1 and p 1 to the alpha r as m r, so (m r)
m 1 to m r are relatively prime down pairwise
m 1 to m r are pairwise relatively prime,
so and now I define n i to be equal to m by
m i, m is the product and I am just this mean
that I am omiting m i in the product, so these
are integers, these are natural numbers i
is from 1 to r and it is obvious that the
gcd of n 1 to n r is 1, So that means I can
write 1 is a combination of n 1 to n r.
So write 1 equal to b 1, n 1 etcetera b r
n r b 1 to b r are integers and therefore
if I take any x, so for every x in x for every
x in G therefore x equal to x power 1 which
is x power b 1 n 1 plus plus plus b r n r
which is same as now these is same as x power
b 1 n 1, x power b 2 n 2, x power b r n r
and these elements I want to call x 1, x 2
and x r, so therefore every element we have
written that is the product of r element x
1 to x r and what is what about x 1 power
(x i) let us take x i power m i these is x
i is x power b i n i and then raise it to
the power m i but that is same thing as x
power b i and n i m i is precisely m so these
is x power m power b i but this x m element
in group G whose order is m therefore this
x power m is identity, so this is also identity
and I am denoting identity by 1 because it
is subgroup of K cross where 1 is identity,
so this is true for every i from 1 to r these
is if you like these is equal to one by Lagrange’s
theorem
So therefore x i power m i is 1 for all 1
to r that is what we have checked but then
remember we are over a field so now look at
this polynomial x (pow) i is fix i from 1
to r and look at the polynomial X power m
i minus 1 these polynomial we know because
its polynomial or a field it has at most n
zeros and I already know that each exercise
is zero of these polynomial each x 1 to x
r is a zero of these polynomial it has at
most m i zeros and I know it has so many zeros
if it has less number of zeros I want to claim
that it has exactly m i zeros because if it
has less number of zeros then G will have
you see here we have proved that every element
of G is a product of x 1 to x r, so if the
number of zeros are of these polynomial is
less than m i than this meaning components
are less, so by cardinality argument (it will)
G will have less number of element, so these
polynomial has exactly 
m i zeros say otherwise would have less number
of zeros otherwise G has less than m element
m is m 1, m 2, m r elements.
So therefore G has these polynomial exactly
has m i zeros also moreover look at this polynomial
X power m i by p i minus 1, these polynomial
has at most m i by p i zeros, so therefore
we can find there exists an element y i and
G such that y i power m i is 1 and y i power
m i by p i is not 1, that is because you see
these has to be the zero of these polynomial,
so it is these and these has more zeros than
these, so therefore I can find an element
y which is zero of these but not zero of these
that means y i I can find so that y i power
m i is 1 and y i power m i by pi is not 1,
But then this means the order of y i so that
is there exists y i in G, i from 1 to r such
that order of y i is precisely m i because
y i power m i is 1 and y i power p i is not
1 therefore order has to be m i and these
is to for every 1 to r and y i is so therefore
order of y 1 to y r these product as earlier
noted that is orders of products of the orders
of y 1 to order of y r and these is m 1 to
m r which is m which is order of G therefore
these elements has to be a generator of a
G.
So that is y 1 to y r, y equal to y 1 to y
r is a generator for G, so G is cyclic, this
is what we wanted to prove, So we have proved
that all finites of groups of the multiplicative
group of a field is finite and we will used
these fact in the next lecture to construct
field arbitrary fields of cardinality p power
n, so we will continue after the break.
