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Section 1.8 Subgroups
Definition 1.8.1 . Subgroup.
A subgroup of a group \(G\) is a subset \(H\subseteq G\) satisfying the following properties.
Identity element.
Closed under group law.
For all \(g_1,g_2\in G\text{,}\) if \(g_1,g_2\in H\text{,}\) then \(g_1g_2\in H\text{.}\) Using logical shorthand:
\begin{equation}
g_1,g_2\in H\implies g_1g_2\in H\text{.}\tag{1.8.1}
\end{equation}
Closed under inverses.
For all \(g\in G\text{,}\) if \(g\in H\text{,}\) then \(g^{-1}\in H\text{.}\) Using logical shorthand:
\begin{equation}
g\in H\implies g^{-1}\in H\text{.}\tag{1.8.2}
\end{equation}
We write \(H\leq G\) to denote that \(H\) is a subgroup of \(G\text{.}\)
Proposition 1.8.3 .
Let \((G, \cdot)\) be a group and suppose \(H\) is a subgroup of \(G\text{.}\)
Restricting the group operation \(\cdot\) to \(H\) defines a binary operation
\begin{align*}
\boldsymbol{\cdot}_H\colon H\times H \amp \rightarrow H\\
(h_1,h_2) \amp \mapsto h_1\boldsymbol{\cdot}_H h_2=h_1\cdot h_2\text{.}
\end{align*}
The pair
\((H, \boldsymbol{\cdot}_H)\) is a group.
Proof.
The only issue here is whether the output of the proposed operation
\(\boldsymbol{\cdot}_H\) actually land in
\(H\text{,}\) and this is guaranteed by axiom (ii) of
Definition 1.8.1 .
Since
\(\boldsymbol{\cdot}_H\) is just the restriction of
\(\cdot\) to the subset
\(H\subseteq G\text{,}\) it is easy to see that it inherits the group axiom properties of
\(\cdot\text{.}\)
For example, since
\(g_1(g_2g_3)=(g_1g_2)g_3\) for
all elements
\(g_1,g_2,g_3\in G\text{,}\) it is certainly true for all
\(g_1,g_2,g_3\in H\text{.}\) Thus
\(\boldsymbol{\cdot}_H\) is associative.
Similarly, since by definition
\(e\in H\) and
\(e\) is the identity with respect to
\(\cdot\text{,}\) it is also is an identity element with respect to
\(\boldsymbol{\cdot}_H\text{.}\) The same argument shows that the inverse of an element of
\(H\) with respect to
\(\cdot\) is also an inverse with respect to
\(\boldsymbol{\cdot}_H\text{;}\) since by definition
\(H\) is closed under the inverse operations, all elements of
\(H\) have inverses with respect to
\(\boldsymbol{\cdot}_H\text{.}\)
Example 1.8.4 . Examples.
Given a group
\(G\) the subsets
\(\{e\}\) and
\(G\) itself are both easily seen to be subgroups.
For all
\(n\in \Z\text{,}\) the set
\(H=\{m\in \Z\mid n\mid m\}\) is a subgroup of
\(\Z\text{.}\)
For all
\(n\in \Z\text{,}\) the set
\(\{nk\mid k\in \Z_{\geq 0}\}\) is
not a subgroup.
For any group
\(G\) and element
\(g\in G\text{,}\) the subset
\(H=\{g^n\mid n\in \Z\}\) is a subgroup of
\(G\text{.}\)
Definition 1.8.5 . Cyclic groups.
Given a group
\(G\) and element
\(g\in G\text{,}\) the subgroup
\(H=\{g^n\mid n\in \Z\}\) is the
cyclic group generated by \(g\) , denoted
\(\angvec{g}\text{.}\)
A group
\(G\) is
cyclic if
\(G=\angvec{g}\) for some
\(g\in G\text{.}\)
Example 1.8.6 . Cyclic groups.
Decide which of the following groups are cyclic. If the example is a family of groups, decide whether all elements of the family are cyclic or not.
\(\displaystyle (\Z/n\Z)^*\)
Solution .
Since
\(\Z=\angvec{1}\text{,}\) \(\Z\) is cyclic.
No
\(D_n\) is cyclic, as a simple order computation shows. Namely, all rotations
\(r^i\) in
\(D_n\) have order at most
\(n\text{,}\) and all reflections
\(r^is\) have order
\(2\text{.}\) It follows that
\(\abs{\angvec{g}}\leq n\) for all
\(g\in D_n\text{.}\) Since
\(\abs{D_n}=2n\text{,}\) we see that there is no
\(g\in D_n\) satisfying
\(D_n=\angvec{g}\text{.}\)
We have
\(\Z/n\Z=\angvec{\overline{1}}\text{,}\) showing that it is cylic.
The group \((\Z/n\Z)^*\) is not always cyclic, the first counterexample being \(n=8\text{.}\) (Show for yourself that this group is cyclic for \(n=2,3,4,5,6,7\text{.}\) ) To see why \((\Z/8\Z)^*\) is not cyclic, we first compute
\begin{equation*}
(\Z/8\Z)^*=\{\overline{1}, \overline{3}, \overline{5}, \overline{7}\} \text{,}
\end{equation*}
and then note that each of these elements has order at most \(2\text{.}\) It follows that \((\Z/8\Z)^*\) is not cyclic.
A cyclic group \(G=\angvec{g}\) is abelian: given \(g_1=g^{n_1}\) and \(g_2=g^{n_2}\text{,}\) we have
\begin{equation*}
g_1g_2=g_2g_1=g^{n_1+n_2}\text{.}
\end{equation*}
As you might have guessed, however, not all abelian groups are cyclic. There are many counterexamples to such a claim, but the smallest one is the group
\begin{equation*}
\Z/2\Z\times\Z/2\Z=\{(0,0),(0,1),(1,0),(1,1)\}\text{.}
\end{equation*}
It is easy to see, using the product group operation defined in
Definition 1.1.5 \(\ord g\leq 2\) for all
\(g\in \Z/2\Z\times \Z/2\Z\text{,}\) and thus that
\(\abs{\angvec{g}}\leq 2\) for all
\(g\in \Z/2\Z\times \Z/2\Z\text{.}\) Since
\(\abs{\Z/2\Z\times \Z/2\Z}=4\text{,}\) we conclude that the group is not cyclic.
This group is important enough to warrant its own name. We call it, and any group that is isomorphic to it, a Klein 4-group . It is easy to see that such a group has a presentation of the form
\begin{equation*}
\angvec{a,b\mid a^2=b^2=e, ab=ba}\text{.}
\end{equation*}
In \(\Z/2\Z\times \Z/2\Z\text{,}\) we can take \(a=(1,0)\) and \(b=(0,1)\text{.}\) We make use of this fact in our official definition of a Klein 4-group.
Specimen 11 . Klein 4-group.
We call a group \(G\) a Klein 4-group if \(G\cong Z/2\Z\times \Z/2\Z\text{.}\) We will use the notation
\begin{equation*}
V_4=\angvec{a,b\mid a^2=b^2=e, ab=ba}
\end{equation*}
to denote a Klein 4-group with generators \(a\) and \(b\text{.}\)
Example 1.8.8 . Subgroups of \(D_n\) .
Find all subgroups of
\(D_3\text{.}\)
Find all cyclic subgroups of
\(D_4\text{.}\) Produce a noncyclic subgroup of
\(D_4\text{.}\)
Definition 1.8.9 . Kernel of homomorphism.
Let \(\phi\colon G\rightarrow H\) be a group homomorphism. The kernel of \(\phi\text{,}\) denoted \(\ker \phi\text{,}\) is the set
\begin{equation}
\ker\phi=\{g\in G\mid \phi(g)=e_H\}\text{.}\tag{1.8.3}
\end{equation}
Theorem 1.8.10 . Kernel and image.
Let \(\phi\colon G\rightarrow H\) be a group homomorphism.
\(\ker\phi\) is a subgroup of
\(G\text{.}\)
\(\im \phi\) is a subgroup of
\(H\text{.}\)
Proof.
Example 1.8.11 . Kernel and image.
Compute \(\ker\phi\) and \(\im \phi\) for the given group homomorphism \(\phi\text{.}\)
Fix
\(m\in \Z\) and define
\(\phi\colon \Z\rightarrow \Z\text{.}\)
\(\phi\colon \Z/12\Z\rightarrow \Z/12\Z\text{,}\) \(\phi(\overline{a})=\overline{3}\overline{a}\)
Let
\(L=\{\ell_1, \ell_2, \ell_3\}\) be the set of three ``diameters" in the regular hexagon
\(P_6\text{.}\) Let
\(\phi\colon D_6\rightarrow S_3\) be the homomorphism associated to the group action of
\(D_6\) on
\(L\text{.}\)
Definition 1.8.12 . Centralizer, normalizer, center.
Let \(A\) be a subset of the group \(G\text{.}\)
The centralizer of \(A\) in \(G\text{,}\) denoted \(C_G(A)\text{,}\) is the set
\begin{equation}
C_G(A)=\{g\in G\mid gag^{-1}=a \text{ for all } a\in A\}=\{g\in G\mid ga=ag \text{ for all } a\in A\}\text{.}\tag{1.8.4}
\end{equation}
The normalizer of \(A\) in \(G\text{,}\) denoted \(N_G(A)\text{,}\) is the set
\begin{equation}
N_G(A)=\{g\in G\mid gAg^{-1}=A\}\text{.}\tag{1.8.5}
\end{equation}
The center of \(G\text{,}\) denoted \(Z(G)\text{,}\) is the set
\begin{equation}
Z(G)=\{g\in G\mid gg'g^{-1}=g' \text{ for all } g'\in G\}=\{g\in G\mid gg'=g'g \text{ for all } g'\in G\}\text{.}\tag{1.8.6}
\end{equation}
\begin{equation*}
\end{equation*}
Theorem 1.8.13 . Centralizer, normalizer, center.
Let \(G\) be a group, and let \(A\) be a subset of \(G\text{.}\)
\(C_G(A)\) is a subgroup of
\(G\text{.}\)
\(N_G(A)\) is a subgroup of
\(G\text{.}\)
\(Z(G)\) is a subgroup of
\(G\text{.}\)
Proof.
Example 1.8.14 . Centralizer, normalizer, center.
Let \(G=S_4\text{.}\)
Let
\(A=\{(1\ 2), (3\ 4)\}\text{.}\) Compute
\(C_{G}(A)\) and
\(N_{G}(A)\text{.}\)
Let
\(B=\{(1\ 2), (1\ 3)\}\text{.}\) Compute
\(C_G(B)\) and
\(N_G(B)\text{.}\)
Compute
\(Z(S_4)\text{.}\)
