# Abstract Algebra/Group Theory/Subgroup

## Subgroups

We are about to witness a universal aspect of mathematics. That is, whenever we have any sort of structure, we ask ourselves: does it admit substructures? In the case of groups, the answer is yes, as we will immediately see.

Definition 1: Let $G$ be a group. Then, if $H\subseteq G$ is a subset of $G$ which is a group in its own right under the same operation as $G$, we call $H$ a subgroup of $G$ and write $H\leq G$.

Example 2: Any group $G$ has at least 2 subgroups; $G$ itself and the trivial group $\{e\}$. These are called the improper and trivial subgroups of $G$, respectively.

Naturally, we would like to have a method of determining whether a given subset of a group is a subgroup. The following two theorems provide this. Since $H$ naturally inherits the associativity property from $G$, we only need to check closure.

Theorem 3: A subset $H$ of a group $G$ is a subgroup if and only if

(i) $H$ is closed under the operation on $G$. That is, if $a,b\in H$, then $ab\in H$,
(ii) $e\in H$,
(iii) $H$ is closed under the taking of inverses. That is, if $a\in H$, then $a^{-1}\in H$.

Proof: The left implication follows directly from the group axioms and the definition of subgroup. For the right implication, we have to verify each group axiom for $H$. Firstly, since $H$ is closed, it is a binary structure, as required, and as mentioned, $H$ inherits associativity from G. In addition, $H$ has the identity element and inverses, so $H$ is a group, and we are done.

There is, however, a more effective method. Each of the three criteria listed above can be condensed into a single one.

Theorem 4: Let $G$ be a group. Then a nonempty subset $H\subseteq G$ is a subgroup if and only if $a,b\in H \,\Rightarrow \, ab^{-1}\in H$.

Proof: Again, the left implication is immediate. For the right implication, we have to verify the (i)-(iii) in the previous theorem. First, assume $a\in H$. Then, letting $b=a$, we obtain $aa^{-1}=e\in H$, taking care of (ii). Now, since $e,a\in H$ we have $ea^{-1}=a^{-1}\in H$ so $H$ is closed under taking of inverses, satisfying (iii). Lastly, assume $a,b\in H$. Then, since $b^{-1}\in H$, we obtain $a(b^{-1})^{-1}=ab\in H$, so $H$ is closed under the operation of $G$, satisfying (i), and we are done.

All right, so now we know how to recognize a subgroup when we are presented with one. Let's take a look at how to find subgroups of a given group. The next theorem essentially solves this problem.

Theorem 5: Let $G$ be a group and $g\in G$. Then the subset $\{g^n\mid n\in \mathbb{Z}\}$ is a subgroup of $G$, denoted $\langle g \rangle$ and called the subgroup generated by $g$. In addition, this is the smallest subgroup containing $g$ in the sense that if $H$ is a subgroup and $g\in H$, then $\langle g \rangle \leq H$.

Proof: First we prove that $\langle g \rangle$ is a subgroup. To see this, note that if $h,k\in \langle g \rangle$, then there exists integers $n,m\in \mathbb{Z}$ such that $h=g^n\,,\,k=g^m$. Then, we observe that $hk^{-1}=g^ng^{-m}=g^{n-m}\in\langle g \rangle$ since $n-m\in \mathbb{Z}$, so $\langle g \rangle$ is a subgroup of $G$, as claimed. To show that it is the smallest subgroup containing $g$, observe that if $H$ is a subgroup containing $g$, then by closure under products and inverses, $g^n\in H$ for all $n\in Z$. In other words, $\langle g\rangle\subseteq H$. Then automatically $\langle g \rangle \leq H$ since $\langle g \rangle$ is a subgroup of $G$.

Theorem 6: Let $H$ and $H^\prime$ be subgroups of a group $G$. Then $H\cap H^\prime$ is also a subgroup of $G$.

Proof: Since both $H$ and $H^\prime$ contain the identity element, their intersection is nonempty. Let $a,b\in H\cap H^\prime$. Then $a,b\in H$ and $a,b\in H^\prime$. Since both $H$ and $H^\prime$ are subgroups, we have $ab^{-1}\in H$ and $ab^{-1}\in H^\prime$. But then $ab^{-1}\in H\cap H^\prime$ (why?). Thus $H\cap H^\prime$ is a subgroup of $G$.

Theorem 6 can easily be generalized to apply for any arbitrary intesection $\bigcap_{i\in I} H_i$ where $H_i$ is a subgroup for every $i$ in an arbitrary index set $I$. The reasoning is identical, and the proof of this generalization is left to the reader to formalize.

Definition 7: Let $G$ be a group and $H$ be a subgroup of $G$. Then $gH=\{gh\mid h\in H\}$ is called a left coset of $H$. The set of all left cosets of $H$ in $G$ is denoted $G/H$. Likewise, $Hg=\{hg\mid h\in H\}$ is called a right coset, and the set of all right cosets of $H$ in $G$ is denoted $H\backslash G$.

Lemma 8: Let $G$ be a group and $H$ be a subgroup of $G$. Then every left coset has the same number of elements.

Proof: Let $g\in G$ and define the function $f\,:\, H \rightarrow gH$ by $h\mapsto gh$. We show that $f$ is a bijection. Firstly, $gh=gh^\prime \,\Rightarrow\, h=h^\prime$ by left cancellation, so $f$ is injective. Secondly, let $h^\prime\in gH$. Then $h^\prime = gh$ for some $h\in H$ and $f(h)=h^\prime$, so $f$ is surjective and a bijection. It follows that $|H|=|gH|$, as was to be shown.

Lemma 9: The relation $\sim$ defined by $a\sim b \,\Leftrightarrow \, a^{-1}b\in H$ is an equivalence relation.

Proof: Reflexivity and symmetry are immediate. For transitivity, let $a\sim b$ and $b\sim c$. Then $a^{-1}b,b^{-1}c\in H$, so $a^{-1}c\in H$ and we are done.

Lemma 10: Let $G$ be a group and $H$ be a subgroup of $G$. Then the left cosets of $H$ partition $G$.

Proof: Note that $aH=bH \Leftrightarrow ah=bh^\prime\,\Leftrightarrow a^{-1}bh^\prime=h \,\Leftrightarrow a^{-1}b\in H$ for some $h,h^\prime\in H$. Since $a\sim b \,\Leftrightarrow \, a^{-1}b\in H$ is an equivalence relation and the equivalence classes are the left cosets of $H$, these automatically partition $G$.

Theorem 11 (Lagrange's theorem): Let $G$ be a finite group and $H$ be a subgroup of $G$. Then $|G|=|G/H||H|$.

Proof: By the previous lemmas, each left coset has the same number of elements $|H|$ and every $g\in G$ is included in a unique left coset $gH$. In other words, $G$ is partitioned by $|G/H|$ left cosets, each contributing an equal number of elements $|H|$. The theorem follows.

Note 12: Each of the previous theorems have analagous versions for right cosets, the proofs of which use identical reasoning. Stating these theorems and writing out their proofs are left as an exercise to the reader.

Corollary 13: Let $G$ be a group and $H$ be a subgroup of $G$. Then right and left cosets of $H$ have the same number of elements.

Proof: Since $H$ is a left and a right coset we immediately have $|gH|=|H|=|Hg^\prime|$ for all $g,g^\prime\in G$.

Corollary 14: Let $G$ be a group and $H$ be a subgroup of $G$. Then the number of left cosets of $H$ in $G$ and the number of right cosets of $H$ in $G$ are equal.

Proof: By Lagrange's theorem and its right coset counterpart, we have $|H||H\backslash G|=|G|=|G/H||H|$. We immediately obtain $|H\backslash G|=|G/H|$, as was to be shown.

Now that we have developed a reasonable body of theory, let us look at our first important family of groups, namely the cyclic groups.

## Problems

Problem 1 (Matrix groups): Show that:

i) The group $GL(n,\mathbb{R})=\{A\in M_n(\mathbb{R})\mid \det(A)\neq 0\}$ of invertible $n\times n$ matrices is a subgroup of $M_n(\mathbb{R})$. This group is called the general linear group of order $n$.
ii) The group $O(n)=\{A\in M_n(\mathbb{R})\mid AA^T=I\}$ of $n\times n$ orthogonal matrices is a subgroup of $GL(N,\mathbb{R})$. This group is called the orthogonal group of order $n$.
iii) The group $SO(n)=\{A\in M_n(\mathbb{R})\mid AA^T=I\,\wedge\, \det(A)=1\}$ is a subgroup of $O(n)$. This group is called the special orthogonal group of roder $n$.
iv) The group $U(n)=\{A\in M_n(\mathbb{C})\mid AA^*=I\}$ of unitary matrices is a subgroup of $GL(n,\mathbb{C})$. This is called the unitary group of order $n$.
v) The group $SU(n)=\{A\in M_n(\mathbb{C})\mid AA^*=I\,\wedge\, \det(A)=1\}$ is a subgroup of $U(n)$. This is called the special unitary group of order $n$.

Problem 2: Show that if $H,K$ are subgroups of $G$, then $H\cup K$ is a subgroup of $G$ if and only if $H\subseteq K$ or $K\subseteq H$.