📂Abstract Algebra

Properties and Proofs of Surplus Types

Theorem

Let $H$ be a subgroup of group $G$. Then, for an element $a$ of group $G$, set $aH= \left\{ ah | h\in H \right\}$ is called the left coset, and $Ha = \left\{ ha | h\in H \right\}$ is called the right coset.

Let’s say $H < G,\enspace a,b \in G,\enspace h \in H$. Then, the following properties are satisfied:

1. $a \in aH$

2. $aH=H \iff a \in H$

3. $aH=bH \iff a \in bH$

4. $aH=bH \ \ \mathrm{or} \ \ aH\cap bH = \varnothing$

5. $aH=bH \iff a^{-1}b \in H$

6. $|aH|=|bH|$

7. $aH=Ha \iff H=aHa^{-1}$

8. $aH \le G \iff a\in H$

Proof

1.

Since $H$ is a subgroup, it contains the identity. $aH$, being the result of operating $a$ with an element of $H$, contains $a=ae \in aH$, and is $a\in aH$.

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2.

• $(\Longrightarrow)$

  Assume $aH=H$. Then, by property 1, $a\in aH=H$ is obtained.

• $(\Longleftarrow)$

  Assume $a\in H$. Since $H$ is a group, it is closed under operation. Therefore, $ah\in H$, and since any element of $aH$ is in $H$, $aH \subset H$ is obtained. Likewise, from $a^{-1}h \in H$,

  $$h=eh=(aa^{-1})h=a(a^{-1}h) \in aH$$

  and since any element of $H$ is in $aH$, $H \subset aH$ is obtained. Since the inclusion holds both ways, $aH=H$ is proven.

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3.

• $(\Longrightarrow)$

  Assuming $aH=bH$, due to property 1, $a\in aH$ and from the assumption, $a \in bH$ is obtained.

• $(\Longleftarrow)$

  Assuming $a \in bH$, it can be expressed as $a=bh$, and the following is obtained:

  $$aH=(bh)H=b(hH)=bH$$

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4.

Let’s say there exists $c$ satisfying $c \in aH \cap bH$. Then, by property 3, $cH=aH$ and $cH=bH$, thus $aH=bH$ is true. Conversely, if there’s no $c$ satisfying $c \in aH \cap bH$, then $aH \cap bH = \varnothing$ is true.

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5.

• $(\Longrightarrow)$

  Assuming $aH=bH$, by property 3, $b \in aH$ and it can be represented as $b=ah$, thus $a^{-1}b=a^{-1}(ah)=h \in H$ is obtained.

• $(\Longleftarrow)$

  Assuming $a^{-1}b \in H$, then $a^{-1}b=h$ and $b=ah$. $b = ah \in aH$ and by property 3, $aH=bH$ is confirmed.

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6.

Consider the function $f : aH \rightarrow bH$. Mapping $ah$ to $bh$, function $f$ becomes a bijection. Thus, $|aH| = |bH|$ is true.

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7.

• $(\Longrightarrow)$

  Assume $aH=Ha$ and let’s consider $aH$ again as $H^{\prime}$. Then, $aH=Ha=H^{\prime}$ is obvious and by substituting $aH$ and $Ha$ respectively into both sides of $H^{\prime}$,

  $$(aH)a^{-1}=(Ha)a^{-1}=H(aa^{-1})=H$$

  thus $aHa^{-1}=H$ is obtained.

• $(\Longleftarrow)$

  Assuming $H=aHa^{-1}$ and using the same method, $Ha=aHa^{-1}a=aH$ is obtained.

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8.

• $(\Longrightarrow)$

  Assuming $aH$ is a subgroup of $G$, then $e \in aH$ is true. Since $e \in H$, $aH \cap eH \ne \varnothing$ is true. By property 4, $aH=eH=H$ is satisfied. By property 2, $a \in H$ is true.

• $(\Longleftarrow)$

  Assuming $a \in H$. By property 2, $aH=H$ is true. Since $H$ is a subgroup of $G$, $aH$ is also a subgroup of $G$.

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See Also

• Normal Subgroup