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Modern Algebra: Groups, Rings, Modules, Fields

Sam Macdonald

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 \(H\neq G\text{.}\)

Proof.

  1. Assume \(H\) is non-empty and for all \(x,y \in H\text{,}\) \(xy^{-1} \in H\text{.}\)
    Since \(H\) is non-empty, there is an \(h \in H\text{.}\) Thus \(hh\inv\in H\text{,}\) and as \(e_G\in H\text{.}\) As \(H\sse G\text{,}\) we see \(e_Gx=x=xe_G\) for any \(x\in H\text{,}\) making \(e_G\) the identity element for \(H\text{.}\)
    For any \(h \in H\text{,}\) \(h^{-1} = eh^{-1} \in H\text{,}\) and so every element of \(H\) has an inverse inside \(H\text{.}\)
    For \(x,y \in H\) we have \(y^{-1} \in H\) and thus \(xy = x(y^{-1})^{-1} \in H\) and hence \(H\) is closed under \(\cdot\text{.}\) This means that the restriction of the group operation of \(G\) to \(H\) is a well-defined group operation. This operation is associative by the axioms for the group \(G\text{.}\) The axioms of a group have now been established for \((H, \cdot)\text{.}\)
  2. 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 \(G\text{.}\)

Example 1.93. Common Subgroups.

  1. \(\{e_G\}\) and \(G\) are subroups of \(G\text{,}\) which we call the trivial subgroups of \(G\text{.}\)
  2. \(\displaystyle \Z\leq\Q\leq\R\leq\C\)
  3. \(\displaystyle \Z^\times\leq\Q^\times\leq\R^\times\leq\C^\times\)
  4. \(A_n\leq S_n\) for all \(n\in\N\text{.}\)
  5. \(\Inn(G)\leq\operatorname{Aut}(G)\text{.}\)
  6. The set of all rotations is a subgroup of \(D_{2n}\text{.}\) Similarly, the set of reflections is also a subgroup.
  7. 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 1.94. Subgroups of \(\Z\).

The only subgroups of \(\Z\) are the sets \(n\Z\text{.}\)
Whilst perusing Example 1.93, you may have noticed the following:
 1 
If you didn’t, don’t worry. I won’t tell anyeone ;)

Definition 1.95. 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*}

Exploration 1.1. Torsion Subgroups.

  1. 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 \(G\text{.}\)
  2. Show that the torision subgroup may not in fact be a subgroup if \(G\) is not assumed to be abelian.

Subsection The Subgroup Lattice

Proof.

  1. Coming soon!
  2. Let \(I\) denote any indexing set and for each \(i \in I\text{,}\) let \(H_i\) be a subgroup of \(G\text{.}\) I claim
    \begin{equation*} H := \bigcap_{i \in I} H_i \end{equation*}
    is a subgroup.
    First, notice \(H\) is nonempty, as \(e_G \in H_i\) for all \(i\text{.}\) If \(x,y \in G\text{,}\) then for each \(i\text{,}\) \(x,y \in H_i\) and hence \(xy^{-1} \in H_i\text{.}\) It follows that \(xy^{-1} \in H\text{,}\) making \(H\) a subgroup by the One Step Subgroup Test.
  3. Coming soon!

Exercise 1.98. No Proper Unions.

No finite group can be expressed as the union of two proper subgroups.
Hint.
Part (3) of Theorem 1.97 may prove useful!

Exercise 1.99. Subgroups not Symmetric.

In Theorem 1.97 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.

Proof.

Let \(\varphi: G \to H\) be a homomorphism of groups.
  1. 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.
  2. 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.
  3. 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 1.101. Inclusions are Homomorphisms.

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

Remark 1.103.

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 1.104. 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*}

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 1.106. Properties of \(Z(G)\).

Let \(G\) be a group.
  1. \(Z(G)\) is abelian.
  2. If \(H\leq G\text{,}\) then \(Z(H)\leq Z(G)\)

Exercise 1.107. 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 1.108. Even Order Groups.

Every group of even order contains an element of order \(2\text{.}\)
 2 
This is a direct result of Theorem 6.4, but proving it is possible with the tools we have.
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