Using Lemma 4.1.6 and the argument preceding it, we have the following.
A subset \(H\) of a group \(G\) is a subgroup of \(G\) if and only if
\(H\) is closed under \(G\)'s operation;
The identity element of \(G\) is in \(H\text{;}\) and
For each \(a\in H\text{,}\) \(a\)'s inverse in \(G\) is contained in \(H\text{.}\)
For each of the following, prove that the given subset \(H\) of group \(G\) is or is not a subgroup of \(G\text{.}\)
\(H=3\Z\text{,}\) \(G=\Z\text{.}\)
\(H=\{0,1,2,3\}\text{,}\) \(G=\Z_6\text{;}\)
\(H=\R^*\text{,}\) \(G=\R\text{;}\)
\(H=\{(0,x,y,z):x,y,z\in \R\}\text{,}\) \(G=\R^4\text{.}\)
Generalizing Part 1 of the above theorem, we have \(n\Z\leq \Z\) for every \(n\in \Z^+\text{.}\)
The proof of this is left as an exercise for the reader.
Consider the group \(\langle F,+\rangle\text{,}\) where \(F\) is the set of all functions from \(\R\) to \(\R\) and \(+\) is pointwise addition. Which of the following are subgroups of \(F\text{?}\)
\(H=\{f\in F: f(5)=0\}\text{;}\)
\(K=\{f\in F: f \text{ is continuous} \}\text{;}\)
\(L=\{f\in F: f \text{ is differentiable} \}\text{.}\)
Are any of \(H\text{,}\) \(K\text{,}\) and \(L\) subgroups of one another?
In fact, we can narrow down the number of facts we need to check to prove a subset \(H\subseteq G\) is a subgroup of \(G\) to only two.
Let \(G\) be a group and \(H\subseteq G\text{.}\) Then \(H\) is a subgroup of \(G\) if
\(H\neq \emptyset\text{;}\) and
For each \(a,b\in H\text{,}\) \(ab^{-1}\in H\text{.}\)
Use the Two-Step Subgroup Test to prove that \(3\Z\) is a subgroup of \(\Z\text{.}\)
Use the Two-Step Subgroup Test to prove that \(SL(n,\R)\) is a subgroup of \(GL(n,\R)\text{.}\)
It is straightforward to prove the following theorem.
If \(H\) is a subgroup of a group \(G\) and \(K\) is a subset of \(H\text{,}\) then \(K\) is a subgroup of \(H\) if and only if it's a subgroup of \(G\text{.}\)
It can be useful to look at how subgroups of a group relate to one another. One way of doing this is to consider subgroup lattices (also known as subgroup diagrams). To draw a subgroup lattice for a group \(G\text{,}\) we list all the subgroups of \(G\text{,}\) writing a subgroup \(K\) below a subgroup \(H\text{,}\) and connecting them with a line, if \(K\) is a subgroup of \(H\) and there is no proper subgroup \(L\) of \(H\) that contains \(K\) as a proper subgroup.
Consider the group \(\Z_8\text{.}\) We will see later that the subgroups of \(\Z_8\) are \(\{0\}\text{,}\) \(\{0,2,4,6\}\text{,}\) \(\{0,4\}\) and \(\Z_8\) itself. So \(\Z_8\) has the following subgroup lattice.
Referring to Example 4.2.4, draw the portion of the subgroup lattice for \(F\) that shows the relationships between itself and its proper subgroups \(H\text{,}\) \(K\text{,}\) and \(L\text{.}\)
Indicate the subgroup relationships between the following groups: \(\Z\text{,}\) \(12\Z\text{,}\) \(\Q^+\text{,}\) \(\R\text{,}\) \(6\Z\text{,}\) \(\R^+\text{,}\) \(3\Z\text{,}\) \(G=\langle \{\pi^n:n\in \Z\},\cdot\,\rangle\) and \(J=\langle \{6^n:n\in \Z\},\cdot\,\rangle .\)
We end with a theorem about homomorphisms and subgroups that leads us to another group invariant.
Let \(G\) and \(G'\) be groups, let \(\phi\) a homomorphism from \(G\) to \(G'\text{,}\) and let \(H\) a subgroup of \(G\text{.}\) Then \(\phi(H)\) is a subgroup of \(G'\text{.}\)
The proof is left as an exercise for the reader.
If \(G\simeq G'\) and \(G\) contains exactly \(n\) subgroups (\(n\in \Z^+\)), then so does \(G'\text{.}\)
This is another way of, for instance, distinguishing between the groups \(\Z_4\) and the Klein 4-group \(\Z_2^2\text{.}\)
By inspection, \(\Z_4\) and \(\Z_2^2\) have, respectively, the following subgroup lattices.
Since \(\Z_4\) contains exactly 3 subgroups and \(\Z_2^2\) exactly 5, we have that \(\Z_4\not\simeq \Z_2^2\text{.}\)