Subgroup

subgroup

A subset $H \subseteq G$ of a Group $(G, ⋆)$ is a subgroup of $G$ if $H$ also forms a group under operation $⋆$ . Namely, it's a subgroup if the restriction $⋆ |_{H} : H \times H \to H$ is a group operation on $H$ .

We usually denote this as $H \leq G$ , read as " $H$ is a subgroup of $G$ ".

Notice that $⋆ |_{H}$ needs to output to $H$ , namely the condition for a Group being closed is required here.

The Subgroup Criterion

Condition for Subgroup, The Subgroup Criterion

Assume $H$ is a nonempty subset of $(G, ⋆)$ which is closed under the binary operation on $G$ and is closed for inverses (so $\forall h, k \in H (h k, h^{- 1} \in H)$ ). Then $H$ is a group under the operation $⋆$ restricted to $H$ (which means the subset $H$ is a subgroup of $G$ ).
(Sometimes we just check $h_{1} h_{2}^{- 1} \in H$ for all $h_{1}, h_{2} \in H$ )
If $H$ is finite, then only closure under the binary operation is needed.

Proof
Note that $⋆ |_{H}$ is already closed by definition. Furthermore:

$⋆ |_{H}$ is associative by the associativity of $⋆$ .
As $H$ is non-empty, then $\exists h \in H$ . Since $H$ is closed under inverses then $h^{- 1} \in H$ . Then:

h ⋆ |_{H} h^{- 1} = h ⋆ h^{- 1} = e_{G} \in H

since $⋆ |_{H}$ is closed. Now for any $h \in H$ :

h ⋆ |_{H} e_{G} = h ⋆ e_{G} = h = e_{G} ⋆ h = e_{G} ⋆ |_{H} h

Thus $e_{G}$ while being the identity of $G$ is also the identity for $H$ .
3. Let $h \in H$ . From the argument above we have $h^{- 1} \in H$ so:

h ⋆ |_{H} h^{- 1} = h ⋆ h^{- 1} = e = h^{- 1} ⋆ h = h^{- 1} ⋆ |_{H} h

Thus then $h^{- 1}$ is also the inverse of $h$ in $H$ .

For the case where $H$ is finite, consider only having closure under the binary operation. Take any $h \in H$ and look at powers of $h$ . Since $H$ is finite, at some point you'll have a repeat (as if it doesn't, then $H$ is infinite):

h, h^{2}, \dots, h^{m} \Rightarrow h^{m} = h^{n}

for some $m, n \in Z$ where $m \neq n$ . For the sake of argument, assume $m < n$ . Then notice that:

1 = h^{n - m}

where $n - m > 0$ . Thus:

h^{- 1} = h^{n - m - 1}

Where $h^{- 1}$ is a product of $h$ 's, so by closure of $H$ under the binary operation, then $h^{- 1} \in H$ .

Aside

For the case where $h k^{- 1} \in H$ for all $h, k \in H$ , since $H$ is non-empty, then $\exists h \in H$ . Thus $h h^{- 1} \in H$ using our given, so $1 \in H$ . Further, take any $k \in H$ . Then $1, h \in H$ so then by our given then $1 h^{- 1} = h^{- 1} \in H$ . For closure on the binary operation, then $h, k \in H$ implies that $k^{- 1} \in H$ so then $h (k^{- 1})^{- 1} = h k \in H$ .

#todo/school : Finish the Subgroup proof at the bottom. 📅 2025-01-16 ✅ 2025-01-16

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Examples

$Z \leq Q, Q \leq R, R \leq C$ so $Z \leq Q \leq R \leq C$
$2 Z = {2 n | n \in Z}$ , and $0 \in 2 Z$ which is closed under addition (even plus even is even, a negative even is an even) is clearly $\leq Z$ .
${2 n + 1 | n \in Z}$ is not a subgroup, since an odd plus an odd is an even (so not closed). Further there's no identity $0 \notin$ the set. It does contain its (negative) inverses though.
Consider $(Q - {0}, \cdot), (R, +)$ are both groups. Clearly $Q - {0} \subseteq R$ , but $(Q - {0}, \cdot) ≰ (R, +)$ due to a difference in operation.
In every group $G$ there are two extreme subgroups:
1. Smallest: ${1_{G}} \leq G$ is the trivial subgroup
2. Largest: $G \leq G$ is the improper subgroup
Remember our Dihedral Groups friend $D_{8} = ⟨ r, s | r^{4} = 1, s^{2} = 1, s r = r^{- 1} s ⟩$ ? What's the smallest subgroup of $D_{8}$ that contains $r$ ? Answer: $⟨ r | r^{4} = 1 ⟩ = {1, r, r^{2}, r^{3}}$ .
1. We call this the subgroup generated by $r$ , denoted $⟨ r ⟩ \leq D_{8}$ .
2. What is $⟨ r, s ⟩ \leq D_{8}$ ? It's the entire set $D_{8}$ !
3. What is $⟨ r^{2}, s ⟩ \leq D_{8}$ ? In this case it's ${1, r^{2}, s, r^{2} s}$ , so it's smaller than we expect!
Here $D_{6} \leq D_{8}$ , because the generators are slightly different. Namely $D_{6} = ⟨ r, s | r^{3} = 1, s^{2} = 1, s r = r^{2} s ⟩$ while $D_{8} = ⟩ r^{'}, s^{'} | (r^{'})^{4} = 1, (s^{'})^{2} = 1, r^{'} s^{'} = s^{'} (r^{'})^{3}$
Consider the symmetric group in $S_{3}$ given as ${1, (1 2) (3)} = ⟨ (1 2) ⟩$ is a subgroup. Furthermore, this subgroup is an Isomorphism to $S_{2}$ .