Subgroups

Section 1.6 Subgroups

Subsection Subgroups

“I have this fear of falling in front of large groups...that’s why I tend not to wear heels.”
―Taylor Swift

Definition 1.91. Subgroup.

A nonempty subset \(H\) of a group \(G\) is called a subgroup, denoted \(H\leq G\text{,}\) if and only if \(H\) is a group under the multiplication law of \(G\text{.}\)

A subgroup \(H\) of a group \(G\) is a called proper subgroup, denoted \(H<G\text{,}\) if and only if \(H\neq G\text{.}\)

Theorem 1.92. Subgroup Tests.

One Step Subgroup Test.
If a subset \(H\) of a group \(G\) is nonempty and satisfies for all \(x,y \in H\text{,}\) \(xy^{-1} \in H\text{,}\) then \(H\) is a subgroup.
Two Step Subgroup Test.
If a subset \(H\) of a group \(G\) is nonempty and closed under multiplication and inversion, then \(H\) is a subgroup.

Proof.

Assume \(H\) is non-empty and for all \(x,y \in H\text{,}\) \(xy^{-1} \in H\text{.}\)

First, we show associativity. Let \(x,y,z\in H\) (not necessarily distinct, as \(H\) might not have three distinct elements). As \(H\leq G\) we have \(x,y,z\in G\text{.}\) As \(G\) is a group is satisfies the associative property, and thus \((xy)z=x(yz)\text{.}\) Hence \(H\) satisfies the associative property as well.

I claim that \(e_G\in H\text{,}\) and further that \(e_G\) is the identity element of \(H\text{.}\) Since \(H\) is non-empty, there exists some \(h \in H\text{.}\) As \(xy^{-1} \in H\) for all \(x,y \in H\text{,}\) setting both \(x\) and \(y=h\) we see \(hh\inv\in H\) Thus \(e_G\in H\text{.}\) As \(H\sse G\text{,}\) we see

\begin{equation*} e_Gx=x=xe_G \end{equation*}

for any \(x\in H\text{,}\) making \(e_G\) the identity element of \(H\text{.}\)

Now I show each element of \(H\) has an inverse in \(H\text{.}\) Let \(h\in H\text{.}\) By setting \(x=e_G\) and \(y=h\text{,}\) we see \(e_Gh\inv\in H\text{.}\) As \(e_G\) is the identity element of \(H\text{,}\) we see \(e_Gh\inv = h\inv\text{,}\) and thus \(h\inv\in H\text{.}\) Hence every element of \(H\) has an inverse contained in \(H\text{.}\)

Finally, we must show that the restriction of the group operation of \(G\) to \(H\) is a well-defined group operation. In other words, given \(x,y\in H\text{,}\) it must be the case that \(xy\in H\text{.}\) Given \(x,y \in H\text{,}\) from the previous paragraph we know \(y^{-1} \in H\text{.}\) Thus \(xy = x(y^{-1})^{-1} \in H\text{,}\) which is what we wanted. This means This operation is associative by the axioms for the group \(G\text{.}\) The axioms of a group have now been established for \((H, \cdot)\text{.}\)
Assume \(H\) is non-empty and closed under multiplication and inversion. Then, for \(x,y\in H\) we have \(y^{-1}\in H\) and \(xy^{-1}\in H\text{.}\) Since the hypothesis of the one-step test is satisfied, \(H\) is a subgroup of \(G\text{.}\)

Remark 1.93.

Note that in our proof of Theorem 1.92 we actually proved something stronger along the way. As we showed \(e_G\) is the (unique) identity element of any subgroup \(H\text{,}\) it is impossible to have a subgroup with a different identity element than its parent group.

Example 1.94. Common Subgroups.

\(\{e_G\}\) and \(G\) are subroups of \(G\text{,}\) which we call the trivial subgroups of \(G\text{.}\)
\(\displaystyle \Z\leq\Q\leq\R\leq\C\)
\(\displaystyle \Z^\times\leq\Q^\times\leq\R^\times\leq\C^\times\)
\(A_n\leq S_n\) for all \(n\in\N\text{.}\)
\(\Inn(G)\leq\operatorname{Aut}(G)\text{.}\)
The set of all rotations is a subgroup of \(D_{2n}\text{.}\) Similarly, the set of reflections is also a subgroup.
Let \(n\in\N\) and define \(n\Z=\{x\in\Z|x=nk\;\text{for some}\;k\in\Z\}\text{.}\) Then \(n\Z\leq\Z\) for all \(n\text{.}\)

In fact, in the case of Item 7, these turn out to be the only subgroups of \(\Z\text{.}\)

Exercise 1.95. Subgroups of \(\Z\).

The only subgroups of \(\Z\) are the sets \(n\Z\text{.}\)

Whilst perusing Example 1.94, you may have noticed the following:

If you didn’t, don’t worry. I won’t tell anyeone.

Exploration 1.1. Torsion Subgroups.

Let \(G\) be an abelian group. Then the set of elements of finite order, which we denote \(T\text{,}\) is a subgroup of \(G\text{,}\) called the torsion subgroup of \(G\text{.}\)
Show that the torision subgroup may not in fact be a subgroup if \(G\) is not assumed to be abelian.

Theorem 1.96. Group Homomorphisms and Subgroups.

Images are Subgroups.
If \(\varphi: G \to H\) is a homomorphism of groups, then the image of \(\varphi\) is a subgroup of \(H\text{.}\)
Kernels are Subgroups.
If \(\varphi: G \to H\) is a homomorphism of groups, then the kernel of \(\varphi\) is a subgroup of \(G\text{.}\)
Preimages are Subgroups.
If \(\varphi:G\to H\) is a group homomorphism and \(K\leq H\) then the preimage of \(K\) is a subgroup of \(G\text{.}\)

Proof.

Let \(\varphi: G \to H\) be a homomorphism of groups.

First, note that since \(\varphi(e_G)=e_H\) by Theorem 1.63, so we know \(\im(\varphi) \ne \emptyset\text{.}\)

Let \(x,y\in \im(\varphi)\text{.}\) Thus \(x = \varphi(a)\) and \(y = \varphi(b)\) for some \(a,b \in G\text{.}\) Hence

\begin{align*} xy^{-1} &= \varphi(a)\varphi(b)^{-1} && \text{Substitution}\\ &= \varphi(ab^{-1}) && \knowl{./knowl/xref/thm-grphomom-preservations.html}{\text{Theorem 1.63}} \end{align*}

As \(xy^{-1}=\varphi(ab^{-1}) \in \im(\varphi)\text{,}\) we see \(\im(\varphi)\leq H\) by the one-step subgroup test.
First, note that since \(\varphi(e_G)=e_H\) by Theorem 1.63, so we know \(\ker(\varphi) \ne \emptyset\text{.}\)

Let \(x,y\in \ker(\varphi)\text{.}\) Then \(\varphi(x)=\varphi(y)=e_G\text{,}\) and we have

\begin{align*} \varphi(xy^{-1}) &= \varphi(x)\varphi(y^{-1}) && \knowl{./knowl/xref/def-ghom.html}{\text{Definition 1.59}}\\ &= \varphi(x)\varphi(y)^{-1} && \knowl{./knowl/xref/thm-grphomom-preservations.html}{\text{Theorem 1.63}}\\ &= \varphi(y)\varphi(y)^{-1} && \text{Substitution}\\ &= e_H && \end{align*}

Thus, if \(x,y\in \ker(\varphi)\) then \(xy^{-1}\in \ker(\varphi)\text{,}\) making \(\ker(\varphi)\leq H\) by the one-step subgroup test.
First notice that \(ke_H = k = e_Hk\) for all \(x\in K\text{,}\) and thus \(e_K = e_H\text{,}\) as and identity elements are unique. As \(e_H=\varphi(e_G)\) by Theorem 1.63, we see \(\varphi\inv(K) \ne \emptyset\text{.}\)

Let \(x,y\in \varphi\inv(K)\text{.}\) Thus \(x,y\in G\) and there exist \(a,b\in K\) such that \(\varphi(x) = a\) and \(\varphi(y) = b\text{.}\) Observe:

\begin{align*} \varphi(xy\inv) &= \varphi(x)\varphi(y\inv) && \knowl{./knowl/xref/def-ghom.html}{\text{Definition 1.59}}\\ &= \varphi(x)\varphi(y)\inv && \knowl{./knowl/xref/thm-grphomom-preservations.html}{\text{Theorem 1.63}}\\ &= ab\inv && \text{Substitution} \end{align*}

As \(K\leq H\text{,}\) we know \(ab\inv\in K\) by the one-step subgroup test. Thus \(\varphi(xy\inv)=ab\inv\in K\text{,}\) and so \(xy\inv\in\varphi\inv(K)\text{.}\) Hence \(\varphi\inv(K)\leq G\) by the one-step subgroup test.

Exercise 1.97. Inclusions are Homomorphisms.

If \(H\) is a subgroup of a group \(G\text{,}\) then the inclusion \(H\into G\) is a group homomorphism.

Definition 1.98. Special Linear Group.

The special linear group with complex entries is defined

\begin{equation*} \SL_n(\C)=\{A \mid A = n\times n \;\text{matrix with entries in}\; \C, \det(A)=1\}. \end{equation*}

Proposition 1.99. \(\SL_n(\C)\leq\GL_n(\C)\).

The special linear group \(\SL_n(\C)\) is a subgroup of the general linear group \(\GL_n(\C)\text{.}\)

Subsection The Subgroup Lattice

Theorem 1.100. Properties of Subgroups.

Transitivity of Subgroups.
If \(H\leq G\) and \(K\leq H\text{,}\) then \(K\leq G\text{.}\)
Intersections of Subgroups.
If \(H_\alpha\) is a subgroup of \(G\) for all \(\alpha\) in an index set \(J\text{,}\) then \(H=\bigcap_{\alpha\in J} H_\alpha\) is a subgroup of \(G\text{.}\)
Unions of Subgroups.
Let \(G\) be a group and \(H\text{,}\) \(H'\) subgroups. The set \(H\cup H'\) is a subgroup of \(G\) if and only if \(H\subseteq H'\) or \(H'\subseteq H\text{.}\)

Proof.

Suppose \(H\leq G\) and \(K\leq H\) for groups \(G,H\text{,}\) and \(K\text{.}\) As \(K\leq H\) we know it is non-empty. For any \(x,y\in K\) we know that \(xy\inv\in K\text{,}\) as \(K\leq H\text{.}\) As \(K\sse G\) as sets, \(K\leq G\) by the One Step Subgroup Test.
Let \(J\) denote any indexing set. For each \(\a \in J\text{,}\) let \(H_\a\) be a subgroup of \(G\text{.}\) I claim

\begin{equation*} H := \bigcap_{\a \in J} H_\a \end{equation*}

is a subgroup of \(G\text{.}\)

First, notice \(H\) is nonempty, as \(e_G \in H_j\) for all \(i\text{.}\) Let \(x,y\in H\text{.}\) Then for each \(\a\) we have \(x,y \in H_\a\text{.}\) As \(H_\a\leq G\) for each \(\a\in J\text{,}\) we have \(xy^{-1} \in H\text{.}\) Thus \(H\) a subgroup by the One Step Subgroup Test.
Coming soon!

Exercise 1.101. No Proper Unions.

No finite group can be expressed as the union of two proper subgroups.

Hint.

Part (3) of Theorem 1.100 may prove useful!

Exercise 1.102. Subgroups not Symmetric.

In Theorem 1.100 we showed that subgroups form a transitve relation. Prove that the relation is reflexive (it can be one sentence) but not symmetric, and thus not an equivalence relation.

Theorem 1.103. Cayley’s Theorem.

Every group is isomorphic to a subgroup of \(S_n\text{.}\)

Remark 1.104.

This is a nearly useless theorem.

Subsection Stuck in the Middle

“Stay in the center, and you will be ready to move in any direction.”
―Alan Watts

Definition 1.105. Center of a Group.

The center of a group \(G\text{,}\) often written \(Z(G)\text{,}\) is the set of elements of \(G\) that commute with every element of \(G\text{.}\) That is,

\begin{equation*} Z(G)=\{x \in G \, | \, xy = yx\,\text{ for all }y \in G\}. \end{equation*}

Proposition 1.106. Center is a Subgroup.

\(\displaystyle Z(G)\leq G\)
\(Z(G)\) is abelian
If \(H\leq G\text{,}\) then \(Z(H)\leq Z(G)\)

Proof.

Let \(G\) be a group. First, notice that since \(xe = x = ex\) for all \(x\in G\text{,}\) we have \(e\in Z(G)\text{,}\) and thus \(Z(G)\ne\emptyset\text{.}\)

Let \(x,y\in Z(G)\text{.}\) I claim \(xy\inv\in Z(G)\text{.}\)

Let \(g\in G\text{,}\) and observe:

\begin{align*} (xy\inv) g &= x(y\inv g) && \\ &= x\left(g\inv (y\inv)\inv\right)\inv && \knowl{./knowl/xref/thm-group-inverses.html}{\text{Theorem 1.21}} (2)\\ &= x(g\inv y)\inv && \knowl{./knowl/xref/thm-group-inverses.html}{\text{Theorem 1.21}} (1) \\ &= x(yg\inv)\inv && y\in Z(G)\\ &= x(gy\inv) && \knowl{./knowl/xref/thm-group-inverses.html}{\text{Theorem 1.21}} (2)\\ &= (gy\inv)x && x\in Z(G)\\ &= g(y\inv x) && \end{align*}

Thus \(Z(G)\leq G\) by the one-step subgroup test.

Exercise 1.107. Properties of \(Z(G)\).

Let \(G\) be a group.

\(Z(G)\) is abelian.
If \(H\leq G\text{,}\) then \(Z(H)\leq Z(G)\)

Exercise 1.108. Only Element of Order \(2\).

Let \(G\) be a group and \(g\in G\text{.}\) If \(g\) is the only element of order \(2\) in \(G\text{,}\) then \(g\in Z(G)\text{.}\)

Exercise 1.109. Even Order Groups.

Every group of even order contains an element of order \(2\text{.}\)

This is a direct result of Theorem 6.4, but proving it is possible with the tools we have.

Summary

A Subgroup is a subset of a group that is also a group under the same operation. The fastest way to show something is a subgroup is with one of the Subgroup Tests.
The Center of a Group is the subgroup of elements of a group that commute with every other element.
³
See: Proposition 1.106.
The Kernel and image of a group homomorphism are subgroups.
⁴
See: Theorem 1.96

Prev Top Next