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I know this question it's proved in prove that $[G:K]$ is finite and $[G:K]=[G:H][H:K]$ but I'd like to understand the demonstration which I was given to

This bellow is the demonstration that I have got from my monitor but I didn't get some steps.

$G=\bigcup_{a \in G}aH$ and $H=\bigcup_{b \in K}bK$

1-Why has he written $G$ and $H$ instead $G:H$ and $H:K$?

Then he mixed $G$ and $H$ and he got:

$G=\bigcup_{i \in A, j \in A}a_ib_jK$

2- Why is it $G$ and not $G:K$ ?

3-and if it's $G:K$ so why $G:H \cup H:K=G:K$?

He tried to prove that the last union (the last $G$) has $(G:H)(H:K)$ distinct elements as written bellow:

-If $i \neq i'$ so $a_ib_jK \neq a_{i'} b_j K$ if they both were equal so $a_ib_j \sim a_{i'} b_j$ (4- where does this equivalence come from?) that's there exists $k \in K$ such that:

$a_{i'}b_j=a_i b_j K$

5-Is this correct because the definition of right equivalence?

The he has written:

$a_{i'}=a_i(b_jKb_j^{-1})$

Then he argued that $b_jKb_j^{-1} \in H$, (6-why?, 7-why does this help?)

he argued that $a_{i'}=a_i(b_jKb_j^{-1})$ is equivalent to:

$a_i H=a_{i'}H$ so there's a contradiction. (8-why contradiction?)

Why does this equivalence hold? why is it not $a_{i'}=a_iH$?

Then he made another step:

-If $j \neq j'$ so we have $a_i b_j K \neq a_i b_{j'} K$ if they were equal so $a_ib_j \sim a_i b_{j'}$ (9- why does he use equivalence and not equality here?) that's there is some $k \in K$ such that $a_ib_{j'}=a_ib_jK$ so:

$b_{j'}=b_jK$

then the same problem again... he argued this is equivalent to:

$b_{j'}K=b_{j}K$

and the proof is complete.

PS: even if this demonstration is correct, is it poorly clean?

Arturo Magidin
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Davi Américo
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2 Answers2

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If $x$ is an element of $G$, and $H$ a subgroup of $G$, then $xH$ is a subset of $G$. By definition, $xH = \{ xh : h \in H\}$. Sets of the forms $xH$ are called left cosets of $H$ in $G$.

$[G : H]$ is the index of $H$ in $G$. It is by definition the number of distinct subsets of $G$ of the form $xH$.

On the other hand, $G/H$ is the set of left cosets of $H$ in $G$. So, $G/H$ is not a subset of $G$, but rather its elements are themselves subsets of $G$. Thus $[G : H]$ is the cardinality of the set $G/H$.

1, 2, 3: The notation

$$G = \bigcup\limits_{a \in G} aH$$

is correct.
This is to say, $G$ is the union of its left cosets.

The statement

$$H = \bigcup\limits_{b \in H} bK$$

is similarly correct. The statement

$$G = \bigcup\limits_{i,j} a_i b_j K$$

is also correct and not a mix up. This is because

$$a_i \Bigg( \bigcup\limits_j b_j K \Bigg) = \bigcup\limits_j a_i b_j K$$

and therefore

$$G = \bigcup\limits_i a_i H = \bigcup\limits_i a_i \Bigg( \bigcup\limits_j b_j K \Bigg) = \bigcup\limits_i \bigcup\limits_j a_i b_j K = \bigcup\limits_{i,j} a_i b_j K.$$

D_S
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  • Why is $G$ the union of left cosets? $|G|=|G:H|=|G:K|$? – Davi Américo Apr 16 '22 at 23:09
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    Why is $G$ equal to the union of left cosets? I hope it is at least clear to you that the union of the left cosets is a subset of $G$, right? – D_S Apr 17 '22 at 22:40
  • you are right, sorry, I get confused about union as if it was ${a_1H,...,a_nH}$ – Davi Américo Apr 17 '22 at 23:15
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    Yes so you see easily that $G$ is equal to the union of its left cosets. Now it is not true that $|G| = [G : H]$ (the number of elements in $G$ is not equal to the number of left cosets of $H$ in $G$), but rather $|G| = [G : H] \times |H|$. – D_S Apr 17 '22 at 23:30
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4- this means $a_ib_jK=a_{i'}b_jK$.

5- I think the K here should change the notation into k, otherwise confusing people with the subgroup K.

6- $b_j$ belongs to H, K belongs to H, then $b_j k b{_j}^{-1}$ belongs to H.

8- $i\not=i'$ so $a_iH\not=a_{i'}H$.

9- this equivalence actually means $a_ib_jk=a_ib_{j'}k$.

ginger cat
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