GATE CSE 1999 | Question: 4

Question

Let $G$ be a finite group and $H$ be a subgroup of $G$. For $a \in G$, define $aH=\left\{ah \mid h \in H\right\}$.

1. Show that $|aH| = |bH|.$

2. Show that for every pair of elements $a, b \in G$, either $aH = bH$ or $aH$ and $bH$ are disjoint.

3. Use the above to argue that the order of $H$ must divide the order of $G.$

Comments

@Deepakk Poonia (Dee) 

simple way ? 

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nice video explanation:

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For Cosets:

https://sites.math.washington.edu/~greenber/Cosets.pdf

https://kconrad.math.uconn.edu/blurbs/grouptheory/coset.pdf

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Is Cosets in GATE 2027 syllabus?

Answer 1

Given :

$\langle G,\star\rangle$ is a finite group and $H$ is a finite subgroup of $G$. For $a \in G$, define $aH = \{ah\mid h \in H\}$

i.e. $aH$ is a left coset.

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a.

The order of a coset is defined as its cardinality.

any $2$ sets have the same cardinality if and only if there is a bijection between them.

So to show any two left cosets have the same cardinality, it suffices to demonstrate a bijection between them.

Let $aH$ and $bH$ are $2$ left cosets of $H$ in $G$ such that

$aH = \{ah\mid h \in H\}$ and $bH = \{bh\mid h \in H\}$ where $a,b \in G$

Let $f$ be a mapping defined from $aH$ to $bH$ such that

$f(ah) = bh\ \forall\ h \in H$

Let $ah_1 =ah_2$ where $h_1,h_2 \in H$

• $\implies a^{-1}ah_1 =a^{-1}ah_2$
• $\implies h_1 =h_2$
• $\implies bh_1 =bh_2$
• $\implies f(ah_1) =f(ah_2)$

$\therefore$ $f$ is well defined.

Let $f(ah_1) =f(ah_2)$ where $h_1,h_2 \in H$

• $\implies bh_1 =bh_2$
• $\implies b^{-1}bh_1 =b^{-1}bh_2$
• $\implies h_1 =h_2$
• $\implies ah_1 =ah_2$

$\therefore$ $f$ is one-one mapping.

Let $y \in bH$ then $y=bh$ where $h \in H$

$\exists ah \in aH$ such that

$f(ah) =hb =y$

• $\implies y$ has a pre-image in $aH$
• $\implies $ Every element of $bH$ has a pre-image in $aH$

$\therefore$ $f$ is an onto mapping.

$\because\ f$ is a well-defined one-one and onto mapping

$\implies f$ is a bijective function

$\implies $  the Cosets $aH$ and $bH$ have the same cardinality.

$\therefore\ |aH| = |bH|$

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b.

For $aH$ and $bH$

Case 1.   $aH$ and $bH$ are disjoint i.e. they have no elements in common. i.e. $(aH \cap bH)=\phi$

Case 2. . $aH=bH$ i.e. all the elements of $aH$ and $bH$ are common(they are identical) i.e. $(aH \cap bH)=(aH \cup bH)$

Case 3. $aH$ and $bH$ have only some common elements and other elements are not common.

i.e We can have (either Case 1. or Case 2. ) and Case 3. should also be satisfied. If Case 1 and Case 2 are TRUE it means Case 3 is false and vice versa. We will now prove that Case 1 and 2 are indeed TRUE.

Let $aH$ and $bH$ be two left cosets of $H$ in $G$ where $a,b \in G$

If we get $aH \cap bH =  \phi$

then Case 1. is satisfied. ($\because$ we assumed it )

Now let us check for Case 2.

Let $aH \cap bH \neq  \phi$ where $a,b \in G$

$\implies$  At least one element of $aH$ must be equal to an element of $bH$

Let $\alpha$ be one such common element

$\implies$ $\alpha \in aH \cap bH$ such that $\alpha =ah_1=bh_2$ for some $h_1,h_2 \in H$. 

$\implies$ $ah_1 = bh_2$

$\implies$ $ah_1h_{1}^{-1} = bh_2h_{1}^{-1}$

$\implies a = b h_2 h_{1}^{-1}$

$\implies a = b h_2 h_{1}^{-1}  \in bH$ ($\because$ $h_1,h_2 \in H \implies h_{1}^{-1},h_2  \in H \implies h_{1}^{−1}h_2 \in H$ by closure property)

$\therefore$ $a \in bH$

So $ a = bh_3$ (where $h_3 = h_2 h_{1}^{-1} \in H$)

$\implies aH = bh_3H$

$\implies aH = bH$ ($\because kH = H$ if $k \in H$)

So, for every pair of elements $a,b \in G$, $aH$ and $bH$ are either disjoint else if they are not disjoint then they will have all the element same.

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c.

From $a.$ and $b.$ we can see that

The cosets partition the entire group $G$ into mutually disjoint subsets i.e.

$G = a_1H \cup a_2H \cup \ldots a_kH $ ($\because$ $2$ left cosets are either identical or disjoint)

$\implies o(G) = o( a_1H) + o(a_2H) + \ldots + o(a_kH) $

$\implies o(G) = o(H)+ o(H) + \ldots+ k\times $ ($\because o(aH)=o(bH) = o(H)$)

$\implies o(G) = k*o(H)$

$\therefore$ the order of $H$ must divide the order of $G$.

This is also known as Lagrange's Theorem.

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NOTE :-

$$kH = H\text{ if }k \in H$$Proof : -

Suppose $k \in H$ and $x \in kH$

$\therefore x = kh$ where $h \in H$

$\implies  xh^{-1} = k $

$\because k \in H \implies xh^{-1} \in H\implies x \in H $(closure property)$\implies kH \subseteq H  \quad \to I$

Let $ x \in H$ then

$x = ex = k k^{-1} x = k(k^{-1} x) \in kH$ ($\because$ $k,x \in H \implies k^{-1},x \in H \implies k^{−1}x \in H$ by closure property)

$\implies H \subseteq kH\quad \to II$

Combining $I$ and $II$ we get $H = kH$

Comments

@Satbir In part a, how can $h_1$ = $h_2$ be true? Is this not a contradiction since H is a set and therefore has unique elements?

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You are missing “Let”

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@Satbir I think the assumption “Let $ah_1 = ah_2$” is wrong because we get the contradiction $h_1 = h_2$. Consider this shorter proof for $|aH| = |bH|$. I will prove that $|aH| = |H|$ for all $a$ belonging to G.

$|aH|$ is not greater than $|H|$ because we are doing element-wise multiplication. Each element $h$ of $H$ is replaced by $ah$. No new element is added. Now suppose $|aH| < |H|$. This is possible only if $|aH|$ has duplicate elements i.e. if $ah_1 = ah_2$ for some $h_1, h_2$ belonging to $H$. But this implies that $h_1 = h_2$ which is a contradiction because sets have unique elements. So only possibility remaining is $|aH| = |H|$ for all $a$ belonging to $G$.

$|aH| = |H|, |bH| = |H|$ for some $a,b$ belonging to $G$. Therefore $|aH| = |bH|$

 

What do you think of this?

Answer 2

Summary for GATE

 

1) Definition of coset:  

  For a subgroup H of a group G and any a in G,  

  aH = { a·h | h in H } is the left coset of H in G.  

  Similarly, Ha = { h·a | h in H } is the right coset.  

  Cosets are shifted copies of H inside G.

 

2) Equal cardinality:  

  For any a in G,  

  the size of the coset aH equals the size of H.  

  All left or right cosets of H have the same number of elements as H.  

 

3) Disjointness or equality:  

  For any a, b in G,  

  either aH = bH or aH ∩ bH = ∅.  

  Thus cosets are either identical or disjoint, and they partition G.  

 

4) Lagrange’s theorem:  

  If G is finite and H is a subgroup,  

  then |G| = [G : H] · |H|.  

  Here [G : H] is the index of H in G (the number of distinct cosets).  

  Also [G : H] = |G| ÷ |H| when G is finite.  

  The set G/H denotes all distinct left cosets of H in G

  G/H is a group only if H is normal.  

If H is normal, then left cosets = right cosets (aH = Ha for all a) and the coset multiplication becomes consistent.  

 

Conclusion 

  Cosets partition G into equal-sized, nonoverlapping subsets,  

  leading to the result that the subgroup’s order divides the group’s order.

Answer 3

Simpler Method

a) Show that $|aH| = |bH|$

Let $H = \{h_1, h_2, \ldots, h_n\}$, then $aH = \{ah_1, ah_2, \ldots, ah_n\}$.

Assume $|aH| \neq |H|$. Then $\exists h_x, h_y \in H$ with $h_x \neq h_y$ such that:

$$ah_x = ah_y$$

By the left cancellation law:

$$h_x = h_y$$

This is a contradiction. $\therefore$ our assumption is wrong and $|aH| = |H|$.

Similarly, $|bH| = |H|$ can be proved, and thus:

$$\boxed{|aH| = |bH| = |H|}$$

b) Show that $aH = bH$ or $aH \cap bH = \emptyset$

Assume $aH \neq bH$ and $aH \cap bH \neq \emptyset$.

Let $p \in (aH \cap bH) \implies p \in aH$ and $p \in bH$.

$$p \in aH \implies p = ah_x$$

$$p \in bH \implies p = bh_y$$

$$\therefore ah_x = bh_y \implies a = bh_y h_x^{-1}$$

Since $H$ is closed under its operation:

$$h_y h_x^{-1} = h_z \quad \text{for some } h_z \in H$$

$$\therefore a = bh_z$$

Since $a = bh_z$ for some $h_z \in H$:

Now, multiply both sides by $H$:

$$aH = (bh_z)H$$

$$aH = b(h_zH)$$

Since $h_z \in H$, by the property of subgroups, we know $h_zH = H$.

$$\therefore aH = bH$$

This is a contradiction. $\therefore$ our assumption is wrong and:

$$\boxed{aH = bH \text{ or } aH \cap bH = \emptyset}$$

c) Lagrange's Theorem

As proved in (a): $|aH| = |bH| = |H|$

Because the cosets partition the group into equal-sized disjoint subsets:

$$\therefore o(H) \mid o(G) $$