Let

(a)

(b)

**Theorem (Frobenius)** - *Berliner Sitzungsberichte, 1895, p. 984
*
If

For the group *G*, let *n* be the largest factor of *g* for which the
Theorem fails. Since the Theorem is clearly true for
*n*=1 and *n*=*g*, it follows that
1<*n*<*g*. Let *N*_{m} denote the number of elements
in *G* whose order divides *m*. Furthermore, let *p* be a prime which
divides *g*. We have

where *N*_{n}^{'} denotes the number of elements in *G* whose order divides
*np* but not *n*. We aim to show that
Since by the
inductive assumption
and therefore
we conclude the
proof if we show that
then *n* divides the difference
*N*_{np}-*N*_{n}^{'}=*N*_{n}.

Let
order of *x* divides *np* but does not divide ;
Write
with (*p*,*s*)=1, that is, *p* and *s* are relatively prime.
Observe that if
then
divides the order of *x*. To show that
divides
we first show that
divides
then
show that *s* divides
(Note also that if
we are done.)

**1.** Let
Then all the
generators of
<*y*_{1}> are in *A*; denote this set by *A*_{1}. Select
Again there are
generators of <*y*_{2}>, all of
them in *A*; call this set *A*_{2}. Note that
,
for if
then
<*y*_{1}>=<*y*>=<*y*_{2}>, a contradiction. Since *A* is finite we have
(disjoint union), and since
divides *A*_{i} for all *i*, we
conclude that
divides
and hence
divides

**2.** (*a*) If
then *x* can be written uniquely as *x*=*yz*,
with *y* and *z* commuting and
and
Indeed, the cyclic
group <*x*> can be written as a direct sum <*x*>=*YZ*, where *Y* is
a subgroup of <*x*> of order
and *Z* is a subgroup of <*x*> of
order *t*. Thus *x*=*yz*, with
and ,
as stated. Furthermore,
since *y* and *z* are in <*x*>, they are both powers of *x*.

As to uniqueness, if *x*=*yz*, with *y* and *z* as above, then
Since
we conclude that
<*y*,*z*>=<*x*>. Thus *y* and *z* are in <*x*> and uniqueness follows from
the internal direct sum decomposition of
<*x*>=<*y*><*z*>. We call *y*
the
*p*-*constituent* of *x*.

(*b*) For
with
let
is the *p*-constituent of
Note that if *y*_{1} and *y*_{2} are two distinct elements of order
in
*A*, then
indeed, this follows from the uniqueness
of the *p*-constituent established in (*a*) above. We can now conclude
that the subsets *C*_{y} partition *A*.

(*c*) Let
Denote by *K*_{y} the centralizer of *y* in *G*.
Clearly
Let
and let
*d*=*gcd*(*r*,*s*), that is, *d* is the greatest
common divisor of *r* and *s*. Since *r*<*g*, by the inductive assumption
there exist a
multiple of *d*, say *cd*, elements in *K*_{y}/<*y*> whose orders divide *d*.
Let
be such a coset of <*y*> in *K*_{y}. We can represent
in the
form *z*<*y*>, with
a divisor of *d*.
[Indeed,
,
and therefore
*z*_{1}^{t}=*y*^{i}, for some *i*. We seek *j* such that *z*=*z*_{1}*y*^{j}
satisfies
1=*z*^{t}=*z*_{1}^{t}*y*^{jt}=*y*^{i}*y*^{jt}=*y*^{i+jt}. Since
such a *j*
exists; take
*j*=-*it*^{-1} (mod
]
Then
Each element of
*K*_{y}/<*y*> whose order divides *d* produces in this manner an element
of *C*_{y}. Thus

(*d*) Consider
the disjoint union taken over all the conjugates
of *y* in *G*. We assert that *s* divides
We have
.
Since *g* is divisible by *s* and *r*, it
follows that *gd* is divisible by *rs*. Hence *s* is a divisor of
Since
*s* and
are relatively prime, *s* must also be a factor of
.
In view of (*b*) above, this shows that *s* divides
This ends the proof.

By making use of the Möbius function we can obtain an expression
for the number of elements whose order is exactly *n* in terms of the
number of elements whose orders divide the divisors *d* of *n*.
Specifically, denote by *f*(*d*) the number of elements whose orders are
exactly *n*. Let *h*(*d*) denote the number of elements whose orders
divide *d*. Clearly we have,
Möbius inversion now
yields
where
is defined to be 0, unless *n*/*d*
is a product of *k* distinct primes, in which case it equals (-1)^{k}.
Frobenius' Theorem informs us that *h*(*d*)=*m* is a multiple of *d*.
We thus conclude as follows:

* *If n divides the order of a group, then the number of elements of *
*order exactly n is equal to*
*. Here the sum is over only *
*those divisors d of n with the property that n/d is a product of *
*any number (k, say) of*
*distinct primes. Furthermore,* *m* *is the number of elements whose *
*orders divide d.*