To prove facts about finite groups, use induction on the order

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To prove facts about finite groups, use induction on the order

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[QUICK DESCRIPTION]
One approach to proving results about finite groups is by using induction
on the order of the group.  The general idea is that one has many constructions
of subgroups, such as the formation of the centre, or of centralizers of elements, or of normalizers of subgroups, to which one can hope to apply an inductive argument.

[PREREQUISITES]
Basic group theory.

[note article incomplete] More examples wanted [/note]

[EXAMPLE]
Cauchy's theorem states that if a prime $p$ divides the order
of a finite group $G$, then $G$ contains an element of order $p$.
We present a proof of Cauchy's theorem via induction on the order
of $G$.  (Many other proofs are possible; a simple proof using an
approach via group actions is 
[[Proving results by letting a group act on a finite set | here]].)

We first consider the decomposition of $G$ into conjugacy classes.
If $g$ is any element of $G$, then the stablizer of $G$ under
the action of $G$ via conjugation is equal to $C_G(g)$, the
centralizer of $g$ in $G$.  Thus the number of conjugates of
$g$ is equal to the index $[G:C_G(g)],$ and so 
if $\{g_1,\ldots,g_n\}$ is a set of conjugacy class representatives
of $G$, 
then 
[maths]
|G| = \sum_{i= 1}^n [G:C_G(g_i)].
[/maths]
Note that $[G:C_G(g)] = 1$ if and only if $C_G(g) = G$,
or equivalently, if and only if $g$ lies in the centre $Z$ of $G$.
Thus if we label our conjugacy class representatives so that
the first $m$ representatives are the elements of the centre,
then we may rewrite the above equation in the form
[maths]
|G| = |Z| + \sum_{i=m+1}^n [G:C_G(g_i)],
[/maths]
where the sum is over the non-central conjugacy class representatives,
i.e. over those conjugacy class representatives whose conjugacy class
contains more than one element.  (This formula is known as the
[[w:conjugacy class | class equation]].)

Now suppose that $p$ does not divide $[G:C_G(g_i)]$ for some 
$i = m+1,\ldots,n$.  Then since $p$ does divide $|G|$, we find
that $p$ divides the order $C_G(g_i)$.  Since $g_i$ is not central
in $G$ (by assumption), we find that $C_G(g_i)$ is a proper subgroup
of $G$.  Thus, by induction, we may conclude that $C_G(g_i)$ contains
an element of order $p$.  Since $C_G(g_i)$ is a subgroup of $G$,
we also get an element of order $p$ in $G$, and so are done.

It remains to consider the case when $p$ divides $[G:C_G(g_i)]$
for all $i  = m+1,\ldots,n$.  Since $p$ divides the order
of $|G|$, we then conclude from the class equation 
that $p$ divides $|Z|$.  Since $Z$ is an abelian group,
this reduces Cauchy's theorem for a general finite group
to the case of a finite abelian group.

Cauchy's theorem for finite abelian groups follows immediately
from the [[w:abelian group | fundamental theorem of finite abelian groups]].
However, we can also prove it directly, using the same strategy
of induction on the order.

Thus suppose that $A$ is a finite abelian group, and that
$p$ is a prime dividing $|A|$.  Let $a$ be any non-identity
element of $A$, and let $B$ be the subgroup of $A$ generated
by $a$.  The order of $B$ is equal to the order of $a$,
which we denote by $r$.  If $p$ divides $r$, then $a^{r/p}$
is an element of $B$, and hence of $A$, of order $p$, and 
we are done.

If $p$ does not divide the order of $B$, then we form
the quotient $A/B$.  (It is here that we use the fact
that $A$ is abelian, so as to be certain that its
subgroup $B$ is normal.)   Since $a$ was chosen to be non-trivial,
the order of $A/B$ is less than the order of $A$,
and since $p$ does not divide the order of $B$ by assumption,
it must divide the order of $A/B$.  By induction,
we conclude that $A/B$ must have an element of order $p$,
say $\overline{a}$.  Let $a' \in A$ be a preimage of $\overline{a}$
under the natural projection $A \rightarrow A/B$ (or, if you prefer,
a representative of the coset $\overline{a}$).
Since the image $\overline{a}$ of $a'$ in $A/B$ has order $p$,
the order of $a'$ must be divisible by $p$.  Thus, if
we let $r'$ denote this order, the element $a^{r'/p}$
of $A$ has order $p$.  This completes the proof of Cauchy's
theorem in the abelian case.

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To prove facts about finite groups, use induction on the order

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