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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 $G.$

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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.

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

$aH = \{ah|h \in H\}$ and $bH = \{bh|h \in H\}$ 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.

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

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. We can't have a case such that $aH$ 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.

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 ) $\implies$ Case 3. so is also satisfied ($\because$ they will have no common elements)

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.

c.

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

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

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

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

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

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

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

This is also known as Lagrange's Theorem.

NOTE :-

$kH = H$ 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 $H = kH$

answered ago by Boss (13.1k points)
selected ago by
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@Arjun Sir, since in the question it is not said to prove can we do like this ?

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No, show that means "to prove". Giving example to show a proof is useless -- you'll get only $0$ mark in IIT/IISc but may be 80% or more in other places.
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In part b, how's the contradiction coming?
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For a part how $a^{−1}h_2 \in H$ is also not clear.
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Updated
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Nice 👍 Such proofs will get 100% in IISc.
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What is the use of Case 3 in part B?
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For 2 sets A and B , 3 type of cases are possible in case of intersection.

Either all elements are same.

No elements are same.

Some elements are same.

The given question only mentions about the first 2 cases.

When I was doing the proof and refering about this topic I was getting confused where is the third case going...what happens if some elements are common in 2 cosets ...thats why i  wrote case 3 and tried to relate it with case 2 and case1 so that i could distinguish them more clearly.
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Case 3 is impossible and proving it is the question. Not sure what's the use of writing it.
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Yes sir, but i wanted to know why its impossible. I just wrote it for my clarity.