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\(\def\Z{\mathbb{Z}} \def\zn{\mathbb{Z}_n} \def\znc{\mathbb{Z}_n^\times} \def\R{\mathbb{R}} \def\Q{\mathbb{Q}} \def\C{\mathbb{C}} \def\N{\mathbb{N}} \def\M{\mathbb{M}} \def\G{\mathcal{G}} \def\0{\mathbf 0} \def\Gdot{\langle G, \cdot\,\rangle} \def\phibar{\overline{\phi}} \DeclareMathOperator{\lcm}{lcm} \DeclareMathOperator{\Ker}{Ker} \def\siml{\sim_L} \def\simr{\sim_R} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} \)

Section4.2Subgroup proofs and lattices

Using Lemma 4.1.6 and the argument preceding it, we have the following.


For each of the following, prove that the given subset \(H\) of group \(G\) is or is not a subgroup of \(G\text{.}\)

  1. \(H=3\Z\text{,}\) \(G=\Z\text{.}\)

  2. \(H=\{0,1,2,3\}\text{,}\) \(G=\Z_6\text{;}\)

  3. \(H=\R^*\text{,}\) \(G=\R\text{;}\)

  4. \(H=\{(0,x,y,z):x,y,z\in \R\}\text{,}\) \(G=\R^4\text{.}\)


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 \(F\text{?}\)

  1. \(H=\{f\in F: f(5)=0\}\text{;}\)

  2. \(K=\{f\in F: f \text{ is continuous} \}\text{;}\)

  3. \(L=\{f\in F: f \text{ is differentiable} \}\text{.}\)

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.


  1. Use the Two-Step Subgroup Test to prove that \(3\Z\) is a subgroup of \(\Z\text{.}\)

  2. 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.

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.

<<SVG image is unavailable, or your browser cannot render it>>


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.

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.

<<SVG image is unavailable, or your browser cannot render it>>

Since \(\Z_4\) contains exactly 3 subgroups and \(\Z_2^2\) exactly 5, we have that \(\Z_4\not\simeq \Z_2^2\text{.}\)