Algebra - Group Theory

Tejas N S
Assistant Professor in Mathematics, NIE First Grade College, Mysuru
Email: nstejas@gmail.com Mobile: +91 9845410469

3. Subgroups


  • Definition 3.1: A non empty subset $H$ of a group $(G,*)$ is said to be subgroup of $G$ if $H$ itself is a group under the same binary operation defined on $G$.

  • Examples:
    (1.) The subset $H=\{1,-1\}$ of $G=\{1,-1,i,-i\}$ is a subgroup under multiplication.
    (2.) $(\mathbb{Z},+)$ is a subgroup of $(\mathbb{Q},+)$.
    (3.) For any integer $n$, the subset $n\mathbb{Z}$ of the group $(\mathbb{Z},+)$ is a subgroup of $(\mathbb{Z},+)$.
    (4.) The subset $H={-1}$ of $G=\{1,-1,i,-i\}$ is not a subgroup under multiplication.

  • Definition 3.2: Let $H$ be a subset of a group $G$. Then the set $H^{-1}$ is defined as $H^{-1}=\{a^{-1}/a \in H\}$. The set $H^{-1}$ is the set of all inverses of the elements of $H$.

  • Theorem 3.1: Let $H$ be a subgroup of group $G$ then identity element of $H$ is same as that of $G$.
    Proof: Let $H$ be a subgroup of $G$. Suppose let $e$ and $e'$ be identities of H and G respectively. To prove: $e=e'$
    Let $a \in H$ $\Rightarrow$ $ae=a$.....(1) (Since $e$ is the identity of $H$)
    Since $a \in H$ $\Rightarrow$ $a \in G$ $\Rightarrow$ $ae'=a$.....(2) (Since $e'$ is the identity of $G$)
    From (1) and (2), we have $ae=ae'$ $\Rightarrow$ $e=e'$ (by left cancellation law) $\blacksquare$

  • Theorem 3.2: Let $H$ and $K$ be subsets of $G$ then $(HK)^{-1}=K^{-1}H^{-1}$.
    Proof: Let $a \in (HK)^{-1}$, then $a=(hk)^{-1}$ where $h \in H$ and $k \in K$. Now, $a=(hk)^{-1}=k^{-1}h^{-1}\in K^{-1}H^{-1}$ $\Rightarrow$ $a \in K^{-1}H^{-1}$
    Hence, $(HK)^{-1}\subseteq K^{-1}H^{-1}$ ......(1)
    Let $b \in K^{-1}H^{-1}$, then $b=k^{-1}h^{-1}$ where $h \in H$ and $k \in K$.
    Now, $b=k^{-1}h^{-1}=(hk)^{-1}\in (HK)^{-1}$ $\Rightarrow$ $b \in (HK)^{-1}$
    Hence, $K^{-1}H^{-1}\subseteq (HK)^{-1}$ ......(2)
    From (1) and (2), we have $(HK)^{-1}=K^{-1}H^{-1}$. $\blacksquare$

  • Theorem 3.3: If $H$ is a subgroup of $G$ then $H^{-1}=H$. Converse may not be true.
    Proof: Let $H$ be a subgroup of a group $G$. Then for $h\in H$ we have $h^{-1} \in H$ (Inverse law). Now $h^{-1} \in H$ $\Rightarrow$ $(h^{-1})^{-1} \in H^{-1}$ $\Rightarrow$ $h \in H^{-1}$. Hence, $H \subseteq H^{-1}$ ......(1)
    Since $H$ is a subgroup of $G$, it contains the inverses of all its elements.
    Hence, $H^{-1}\subseteq H$ ......(2)
    From (1) and (2), we have $H^{-1}=H$
    Converse of the above theorem may not be true: i.e., if $H$ is a subset of $G$ and $H^{-1}=H$, then $H$ may not be a subgroup of $G$.
    For example: $G=\{-1,1\}$ is a group under multiplication. Consider $H=\{-1\}$ and so $H^{-1}=\{-1\}$. But $H^{-1}$ is not a subgroup of $G$.(Why!) $\blacksquare$

  • Theorem 3.4: (Necessary and sufficient condition for a non empty subset to be a subgroup of a group) Let $H$ be a non empty subset of a group $G$. Then $H$ is a subgroup of $G$ if and only if $\forall$ $a,b \in H$ $\Rightarrow$ $ab^{-1} \in H$.
    Proof: Necessary condition: Let $H$ be a subgroup of $G$. Then $H$ itself is a group. So $b \in H$ $\Rightarrow$ $b^{-1} \in H$ (by inverse law) and $a,b^{-1} \in H$ $\Rightarrow$ $ab^{-1} \in H$ (by closure law)
    Sufficient condition: Conversely, Suppose let $H$ is a non empty subset of $G$ with $a,b \in H$ $\Rightarrow$ $ab^{-1} \in H$. To prove: $H$ is a subgroup of $G$. i.e., to prove $H$ is a group.
    (1.) Existence of identity: Let $a,a \in H$ $\Rightarrow$ $aa^{-1} \in H$ $\Rightarrow$ $e \in H$.
    (2.) Existence of inverse: Let $e,a \in H$ $\Rightarrow$ $ea^{-1} \in H$ $\Rightarrow$ $a^{-1} \in H$.
    (3.) Closure law: Let $b \in H$ $\Rightarrow$ $b^{-1} \in H$.
    Now let $a,b^{-1} \in H$ $\Rightarrow$ $a(b^{-1})^{-1} \in H$ $\Rightarrow$ $ab \in H$.
    (4.) Associative law: Since associative law holds in $G$, it also holds in $H$ as $H$ is a subset of $G$.
    $\therefore$ $H$ is a group $\Rightarrow$ $H$ is a subgroup of $G$. $\blacksquare$

  • Theorem 3.5: A non empty subset $H$ of a group $G$ is a subgroup if and only if $HH^{-1}=H$.
    Proof: Let $H$ be a subgroup of $G$.
    To prove: $HH^{-1}=H$
    Let $a \in HH^{-1}$. Then $a=hk^{-1}$ where $h,k \in H$
    For $k \in H$ $\Rightarrow$ $k^{-1} \in H$ (by inverse law) and for $h,k^{-1} \in H$ $\Rightarrow$ $hk^{-1} \in H$ and so $a \in H$
    Hence, $HH^{-1}\subseteq H$ ......(1)
    Let $b \in H$ $\Rightarrow$ $be^{-1} \in HH^{-1}$ $\Rightarrow$ $be \in HH^{-1}$ $\Rightarrow$ $b \in HH^{-1}$
    Hence, $H\subseteq HH^{-1}$ ......(2)
    From (1) and (2), we have $HH^{-1}=H$
    Conversely, let $H$ be a non empty subset of $G$ such that $HH^{-1}=H$.
    To prove: $H$ is a subgroup of $G$.
    Let $h,k \in H$ $\Rightarrow$ $hk^{-1} \in HH^{-1}$ $\Rightarrow$ $hk^{-1} \in H$
    $\Rightarrow$ $H$ is a subgroup of $G$. (By theorem 3.4) $\blacksquare$

  • Theorem 3.6: If $H$ and $K$ are subgroups of $G$ then $HK$ is a subgroup of $G$ if and only if $HK=KH$.
    Proof: Let $H$ and $K$ are subgroups of $G$.
    Let $HK$ be a subgroup of $G$. To prove: $HK=KH$
    Since $H,K$ and $HK$ are subgroups of $G$, we have $H=H^{-1}$, $K=K^{-1}$ and $HK=(HK)^{-1}$ (by theorem 3.3)
    Now, $HK=(HK)^{-1}=K^{-1}H^{-1}=KH$
    Hence, $HK=KH$
    Conversely: Suppose if $HK=KH$.
    To Prove: $HK$ is a subgroup of $G$. It is enough if we prove $(HK)(HK)^{-1}=HK$.
    Now, $(HK)(HK)^{-1}=(HK)(K^{-1}H^{-1})$
    $=H(KK^{-1})H^{-1}$
    $=(HK)H^{-1}$
    $=(KH)H^{-1}$
    $=K(HH^{-1})$
    $=KH=HK$
    Hence, $(HK)(HK)^{-1}=HK$ $\Rightarrow$ $HK$ is a subgroup of $G$. $\blacksquare$

  • Theorem 3.7: Intersection of two subgroups of a group is a subgroup.
    Proof: Let $H_1$ and $H_2$ be two subgroups of $G$.
    To prove: $H_1\cap H_2$ is a subgroup of $G$. It is enough if we prove that $a,b \in H_1 \cap H_2$ $\Rightarrow$ $ab^{-1} \in H_1 \cap H_2$. (By theorem 3.4)
    Since $H_1$ and $H_2$ are subgroups of $G$ $\Rightarrow$ $H_1$ and $H_2$ are non empty subsets of $G$ $\Rightarrow$ $H_1 \cap H_2$ is also a non empty subset of $G$.
    Let $a,b \in H_1 \cap H_2$ $\Rightarrow$ $a,b \in H_1$ and $a,b \in H_2$
    Since $H_1$ is a subgroup of $G$ we have $a,b \in H_1$ $\Rightarrow$ $ab^{-1} \in H_1$
    Since $H_2$ is a subgroup of $G$ we have $a,b \in H_2$ $\Rightarrow$ $ab^{-1} \in H_2$
    Thus $ab^{-1} \in H_1$ and $ab^{-1} \in H_2$ $\Rightarrow$ $ab^{-1} \in H_1 \cap H_2$ $\blacksquare$

  • Problem 3.1: Is $\mathbb{Z}$ a subgroup of $(\mathbb{Q},+)$?
    Solution: $\mathbb{Z}$ is a non empty subset of $\mathbb{Q}$ and $(\mathbb{Z},+)$ is an group. So $(\mathbb{Z},+)$ a subgroup of $(\mathbb{Q},+)$. $\spadesuit$

  • Problem 3.2: Is $\mathbb{Z}$ a subgroup of $(\mathbb{Q},\times)$?
    Solution: $\mathbb{Z}$ is a non empty subset of $\mathbb{Q}$ and $(\mathbb{Z},\times)$ is not a group. So $(\mathbb{Z},\times)$ is not a subgroup of $(\mathbb{Q},\times)$. $\spadesuit$

  • Problem 3.3: Define center of a group $G$ and show that its a subgroup of $G$.
    Definition 3.3: The center of a group $G$ is the set of elements of $G$ which commute with every element of $G$ i.e., $Z=\{z \in G/zx=xz$ $\forall$ $x \in G\}$.
    Solution: $Z=\{z \in G/zx=xz$ $\forall$ $x \in G\}$
    Clearly $Z$ is a non empty subset of $G$.
    To prove: $Z$ is a subgroup of $G$. i.e., $a, b \in Z \Rightarrow ab^{-1 } \in Z$
    Let $a,b \in Z$ $\Rightarrow$ $ax=xa$ and $bx=xb$ $\forall$ $x \in G$.
    Consider, $xa=ax$
    $\Rightarrow$ $xa=a(b^{-1}b)x$
    $\Rightarrow$ $xa=(ab^{-1})(bx)$
    $\Rightarrow$ $xa=(ab^{-1})(xb)$
    $\Rightarrow$ $(xa)b^{-1}=(ab^{-1})(xb)b^{-1}$
    $\Rightarrow$ $x(ab^{-1})=(ab^{-1})x(bb^{-1})$
    $\Rightarrow$ $x(ab^{-1})=(ab^{-1})x$
    $\Rightarrow$ $ab^{-1} \in Z$ $\spadesuit$

  • Problem 3.4: Define Normalizer of a group $G$ and show that its a subgroup of $G$.
    Definition 3.4: Let $G$ be a group and $a \in G$, then the normalizer of $a$ in $G$ is the set of all those elements of $G$ which commutes with $a$ and is denoted by $N_a$ i.e., $N_a=\{x \in G/ ax=xa \}$
    Solution: $N_a=\{x \in G/ ax=xa \}$
    Clearly $N_a$ is a non empty subset of $G$.
    Let $x,y \in N_a$ then $ax=xa$ and $ay=ya$
    To prove: $N_a$ is a subgroup of $G$, it is enough to prove that $a(xy^{-1})=(xy^{-1})a$.
    Now consider, $ya=ay$
    $\Rightarrow$ $y^{-1}(ya)y^{-1}=y^{-1}(ay)y^{-1}$
    $\Rightarrow$ $(y^{-1}y)ay^{-1}=y^{-1}a(yy^{-1})$
    $\Rightarrow$ $ay^{-1}=y^{-1}a$
    $\Rightarrow$ $y^{-1} \in N_a$
    Now consider, $(xy^{-1})a=x(y^{-1}a)=x(ay^{-1})=(xa)y^{-1}=(ax)y^{-1}=a(xy^{-1})$
    $\therefore$ $(xy^{-1})a=a(xy^{-1})$ $\Rightarrow$ $xy^{-1} \in N_a$ $\spadesuit$

  • Problem 3.5: If $H$ is a subgroup of $G$ and $g \in G$, then prove that $\{ghg^{-1}/h \in H\}$ is a subgroup of $G$.
    Solution: For $g \in G$, let $gHg^{-1}=\{ghg^{-1}/h \in H\}$
    Let $x,y \in gHg^{-1}$ $\Rightarrow$ $x=gh_1g^{-1}$ and $y=gh_2g^{-1}$ for some $h_1,h_2 \in H$
    Since $H$ is a subgroup of $G$, we have $h_1,h_2 \in H$ $\Rightarrow$ $h_1h_2^{-1} \in H$
    Consider, $xy^{-1}=gh_1g^{-1}(gh_2g^{-1})^{-1}=gh_1g^{-1}(gh_2^{-1}g^{-1})=gh_1(g^{-1}g)h_2^{-1}g^{-1}=gh_1h_2^{-1}g^{-1}$
    Since $h_1h_2^{-1}\in H$, we have $gh_1h_2^{-1}g^{-1} \in gHg^{-1}$ $\Rightarrow$ $xy^{-1} \in gHg^{-1}$ $\Rightarrow gHg^{-1}$ is a subgroup of $G$. $\spadesuit$

  • Problem 3.6: Show that union of two subgroups is not necessarily a subgroup.
    Solution: $(\mathbb{Z}_6,\oplus_6)$ is a group.
    Consider $H_1=\{0,3\}$ and $H_2=\{0,2,4\}$. Then $H_1$ and $H_2$ are subgroups of $\mathbb{Z}_6$. (How!)
    But $H_1 \cup H_2=\{0,2,3,4\}$ is not a subgroup of $\mathbb{Z}_6$. For $2 \oplus_6 3=5 \notin H_1 \cup H_2$ and so Closure law does not hold in $H_1\cup H_2$. $\spadesuit$