This approach goes along elementary ideas. But the proof using the power group homomorphism and the size estimate of the kernel and the image using number of the polynomial roots is much more elegant and simple.

The number $m$ of elements of a subgroup $H$ divides the number $n$ of elements of a group $G$. It simply follows by freely acting by translation with the subgroup on the group (number of the orbits times the number of elements in the subgroup gives the number of elements in the group). The order $k$ of any element is the size of the subgroup generated by that element, therefore $k$ divides $n$. In particular, for any $g$ in $G$, $g^n=1$ (well known!).

*Lemma 1.* Let $a$ be an element of order $u$ and $b$ an element of order $v$ in $G$ (commutative). If nothing but $1$ divides both $u$ and $v$, then:

(I) the intersection of the two groups generated by $a$ and $b$ contains $1$ and nothing more.

(II) the order of the product $a b$ equals the product $u v.$

*Proof:* (I) the intersection being a subgroup of both cyclic groups, the number of the elements of the intersection should divide both $u$ and $v$.

(II) $(a b)^{u v}= 1 = (a^u)^v (b^v)^u$, hence the order $l$ of $a b$ is a divisor of $u v$. Also $a^l b^l =1$, hence $a^l=b^{-l}$. From (I) above we get $a^l=1$ and $b^l=1$; hence $u$ and $v$ divide $l$. Therefore $l = u v$.

*Lemma 2.* If $a$ has order $u$ and $u = w t$ then $c=a^w$ has order $t$ . (obvious).

Using the above results I want to show that:

*Prop 1.* In a commutative group that is not cyclic there are two distinct cyclic subgroups of the same order (with some elements in both).

*Corollary.* In a commutative group that is not cyclic, for some integer $w$ there are $w+1$ distinct elements $h$ with the property $h^w=1$.

*Proof:* Suppose $a$ is chosen in $G$ such it has the highest possible order $m$ and $m < n$ (the number of elements in $G$). Take $b\neq a^i$, i.e. $b$ is not in the group generated by $a$; suppose $b$ has order $k$. We show that $k$ divides $m$. If $k$ has a divisor $p^j$ ($p$ prime) that does not divide $m$ (this means either $p$ does not divide $m$ or the power $j$ is higher than the power of $p$ dividing $m$), using the results above we can build a new element having order $w >= p m$ (first, using lemma 2, raise $a$ to a power that takes out $p$ from its order, second raise $b$ to a power that takes out all primes except $p$ from its order and finally, using lemma 1(II), multiply the two elements), which contradicts how $a$ was chosen.

Therefore $k$ divides $m$ and $a_0=a^{m/k}$ generates a group of order $k$ as $b$ does. Moreover $b$ is not in that subgroup. These are the subgroups looked for.

*Prop 2.* Suppose $G$ is a finite group (with respect to multiplication) inside a field. Then $G$ is cyclic.

If $G$ is not cyclic, using Corollary above we get an integer $w$ for which $X^w-1$ has $w+1$ roots. Since we are in a field, this is impossible, as for every root $r$ of a polynomial we get a factor $(X-r)$ of that polynomial. And a polynomial of degree $d$ cannot have more than $d$ factors.

wantto introduce Euler's $\varphi$ function in a number theory course? As a number theorist, I would defend introducing it even in a pure algebra course, but in a number theory course it seems almost mandatory. $\endgroup$ – Pete L. Clark Feb 8 '11 at 10:02