Lagrange's Theorem.
If $H$ is a subgroup of a finite
group $G,$ then the order of $H$
is a divisor of the order of $G.$
Note.
(Page 83)
Since $\abs{S_3}=3!=6,$
any subgroup of $S_3$ must
have order $1,2,3,$ or $6;$
$S_3$ cannot have subgroups of
order $4$ or $5.$ A group of
order $7$ can have only the two
obvious subgroups: $\{e\},$
of order $1;$ and the group itself,
of order $7.$
Definition.
(Page 84)
The integer $k$ appearing
in the proof of Lagrange's Theorem
is called the
index of $H$ in $G.$
This index will be denoted
$[G:H].$
Thus,
$[G:H]$
is the number of right cosets of
$H$ in $G,$ and
\[
\abs{G}=\abs{H}\cdot[G:H].
\]
Notice that this equation shows
that $[G:H],$ as well as $\abs{H},$
is a divisor of $\abs{G}.$
Corollary.
(Page 84)
If $G$ is a finite group and $a\in G,$
then the order of $a$ is a divisor
of the order of $G.$
Corollary.
(Page 84)
A group $G$ of prime order contains
no subgroup other than $\{e\}$
and $G.$
Corollary.
(Page 84)
Each group $G$ of prime order is cyclic,
generated by any one of its nonidentity
elements.
Example 17.1.
Groups of prime-squared order
need not be cyclic. For example,
$\Z_p\times\Z_p$ has order $p^2$
but is not cyclic because it has no
element of order $p^2;$ each of its
nonidentity elements has order $p.$
Example 17.2.
Lagrange's Theorem simplifies
the problem of determining subgroups
of a finite group. Let $G$ denote the
Symmetry Group of the Square.
Aside from $\{\mu_1\}$ and the whole
group, any subgroup must have order
$2$ or $4.$
Subgroups of order $2$ are easy to
determine —
each contains the identity
together with an element of order
$2.$ (These correspond to appearances
of $\mu_1$ on the diagonal of the
Cayley table for symmetries of a square.)
In this case, the following subgroups
all have order $2.$
\[
\gen{\mu_3},
\gen{\mu_5},
\gen{\mu_6},
\gen{\mu_7},
\gen{\mu_8}
\]
This leaves only order $4,$
and careful inspection will reveal
three subgroups of this order:
\[
\gen{\mu_2},
\gen{\mu_5,\mu_6},
\gen{\mu_7,\mu_8}.
\]
Definition.
(Page 84)
Figure 17.1
shows the subgroups of $G$
from Example N 17.2
and the inclusion relations between them.
This is an example of a
subgroup lattice.
Such a figure is constructed as follows.
If $A$ and $B$ are subgroups of $G$
with $A\subsetneq B,$ and there is no
subgroup $C$ such that
$A\subsetneq C\subsetneq B,$
then $B$ appears above $A$ and a segment
is drawn connecting $A$ and $B.$
Note.
(Page 85)
Lagrange's theorem does not
say that if $n$ is a divisor of the
order of $G,$ then $G$ has a subgroup
of order $n.$ That would be false.
For example, there is a group of order
$12$ having no subgroup of order $6.$
However, Theorem 17.1 shows
that if $G$ is a cyclic finite
group of order $n,$ then it does
have a subgroup of every order
dividing $n.$
Guevara Note.
(Page 85)
A cyclic group need not be finite.
See pages 72-73.
Theorem 17.1.
Let $G$ be a cyclic group of (finite)
order $n,$ with $G=\gen{a}$
$=\{e,a,a^2,\ldots,a^{n-1}\}.$
-
Every subgroup of $G$ is cyclic.
-
If $1\le k\lt n,$ then $a^k$
generates a subgroup of order
$n/(k,n),$ where $(k,n)$
is the greatest common divisor
of $k$ and $n.$
-
For each positive divisor $d$
of $n,$ $G$ has exactly one
subgroup of order $d.$
Example 17.3.
The positive divisors of $12$ are
$1,2,3,4,6,12.$ Thus, every cyclic
group of order $12$ has exactly one
(cyclic) subgroup of each of these
orders. Figure 17.2 illustrates
this by showing the subgroup lattice
of $\Z_{12}.$ The notation $a^k$ in
Theorem 17.1 becomes $k[a]$
in this example.
Alex Notes