A subset $H$ of a group $G$
is a
subgroup
of $G$ if $H$
is itself a group with respect
to the operation on $G.$
The definition of subgroup implies
that if $H$ is a subgroup of $G$
then $H$ must be closed with
respect to the operation $*$
on $G.$ That is, if $a,b\in H$
then $a*b\in H.$
In particular, for each $a\in H,$
\[
a*a\in H.
\]
The set of odd integers is not closed
with respect to addition, since an odd
integer plus an odd integer is an even
integer. Therefore, the set of odd
integers with addition is not a group,
nor a subgroup of the group of integers.
The set of even integers is closed
with respect to addition since an
even integer plus an even integer
is an even integer. It can be
further verified that the set of
even integers is a group, and a
subgroup of the group of integers.
Some examples of subgroups:
-
The group of integers with addition
is a subgroup of the group of real
numbers with addition.
-
With multiplication, $\{1,-1\}$
is a subgroup of the group of
nonzero real numbers.
-
Any group is a subgroup of itself.
-
If $e$ is the identity of a group $G$
then $\{e\}$ is a subgroup of $G.$
Let $G$ be a group with operation $*$
and let $H$ be a subgroup of $G.$
-
If $f$ is the identify of $H$
and $e$ is the identity of $G$
then $f=e.$
-
If $a\in H$ then the inverse of $a$ in $H$ is
the same as the inverse of $a$ in $G.$
Let $G$ be a group with operation $*$ and
let $H$ be a subset $G.$ Then $H$ is a subgroup
of $G$ iff
-
$H$ is nonempty,
-
if $a\in H$ and $b\in H$ then $a*b\in H,$
-
if $a\in H$ then $a^{-1}\in H.$
(page 43, problem 7.22)
Let $G$ be a group with operation $*$ and
let $H$ be a subset $G.$ Then $H$ is a subgroup
of $G$ iff
-
$H$ is nonempty,
-
if $a,b\in H$ then $a*b^{-1}\in H,$
(page 75, problem 14.35)
Let $G$ be a group with operation $*$ and
let $H$ be a finite subset $G.$ Then $H$ is a subgroup
of $G$ iff
-
$H$ is nonempty,
-
if $a,b\in H$ then $a*b\in H.$
That is, condition (c) of
If $k$ is an integer then
the integral multiples of $k$
is a subgroup of $\Z$ with addition,
since this
group of the integral multiples of $k$
satisfies the conditions of
Let $H=\{(1), (1\ 2\ 3), (1\ 3\ 2)\}.$
Then $H$ is a subgroup of $S_3.$
Draw the Cayley table and use
For any $n,$ a
transposition
is a 2-cycle in $S_n.$
Every element of $S_n$ is a transposition
or a product of transpositions.
$(1\ 2\ 3)=(1\ 3)(1\ 2)$
A permutation is
even
or
odd
if it can be written as a product
of an even or odd number of
transpositions, respectively.
The number of transpositions needed
to factor a permutation is necessarily
either even or odd depending only on
the given permutation.
A permutation can be written as a product
of transpositions in more than one way, such as
$(1\ 2\ 3)$
$=(1\ 3)(1\ 3)$
$=(2\ 3)(1\ 2)(1\ 3)(2\ 3).$
Although their number of transpositions is different,
both factorizations have an even number of transpositions.
The preceding theorem ensures this will always
hold.
The set of all even permutations in $S_n$
is called
the alternating group of degree $n$
denoted by
$A_n.$
$A_n$ is a subgroup of $S_n$ for each
$n\ge2.$
The order of $A_n$ is $\frac{1}{2}(n!).$
The group $H$ in
Find $A_4.$
Let $G$ be a permutation group on $S,$
$T\subseteq S,$
and
\[
G_T=\{\alpha\in G\mid\alpha(t)=t,\ \forall t\in T\}.
\]
We say that
the elements of $G_T$ leave $T$
elementwise invariant.
Let $S=\{1,2,3,4\},$
$G=\Sym(S)=S_4,$
and $T=\{1,2\}.$
Then
\[
\begin{align*}
G_T&=\{(1)(2)(3)(4), (1)(2)(3\ 4)\}\\
&=\{(1),(3\ 4)\}
\end{align*}
\]
Let $G$ be a permutation group on $S,$
$T\subseteq S,$
and
\[
G_{(T)}=\{\alpha\in G\mid\alpha(T)=T\},
\]
where $\alpha(T)=\{\alpha(t)\mid t\in T\}.$
Thus, if $\alpha\in G_{(T)}$
then $\alpha$ may permute the elements
of $T$ among themselves, but it sends
no element of $T$ outside $T.$
We say that
the elements of $G_{(T)}$
leave $T$ invariant.
With $S,G$ and $T$ as in
If $G$ is a permutation group on $S$
and $T\subseteq S,$
then $G_T$ and $G_{(T)}$
are subgroups of $G.$
Also, $G_T$ is a subgroup of $G_{(T)}.$
That is,
\[
G_T\subseteq G_{(T)}\subseteq G.
\]
The
Alex Notes