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Lagrange Theorem states that the order (number of elements) of every subgroup of G is divisible by the order of G itself. It's easy to understand, yet compelling.

The issue is that I really don't know much about Group Theory, but I've been able to help myself somehow with the Internet. However, I've become stuck with a very simple Wikipedia proof of Lagrange Theorem. It has some vocabulary that I had to look up and learn, so maybe the issue is that I didn't comprehend some words...I don't know.

I'm thinking of quoting each sentence of the proof and explaining my understanding of it. I figure that in that way someone might find where I got something wrong. I would really appreciate any help/thoughts! If there is something that isn't clear from my explanation I would be totally happy to clarify.


"Proof of Lagrange's theorem

This can be shown using the concept of left cosets of H in G."

  • H is a subgroup of G. A left coset of H is defined as gH, where g is an element of G.

"The left cosets are the equivalence classes of a certain equivalence relation on G and therefore form a partition of G."

  • Since it says 'therefore', I believe that the significant part in the latter sentence is that the left cosets of H are a partition of G. Which would be true since for each g and H we choose, the elements of gH (because of the definition of a group) would be in G.

"Specifically, x and y in G are related if and only if there exists h in H such that x = yh."

  • If this statement is more than a definition then I don't get it. I would understand that it is defining what conditions must a pair of elements have in order to be related.

"If we can show that all cosets of H have the same number of elements, then each coset of H has precisely |H| elements."

  • If a coset has an element twice, then it would not have |H| elements, right? Or are they claiming that there exists no such thing as a coset with repeated elements?

"We are then done since the order of H times the number of cosets is equal to the number of elements in G, thereby proving that the order of H divides the order of G."

  • Again, I'm confused. Isn't a coset defined as gH (g being an element of G)? So, since there are |G| elements we can choose from, there should be |G| left cosets of H...right? And for |G| times |H| to give |G|, |G |H| would have to be 1. This sentence also made me become sure that I likely got some definitions wrong.

"Now, if aH and bH are two left cosets of H, we can define a map f : aH → bH by setting f(x) = ba−1x. This map is bijective because its inverse is given by f−1(y) = ab−1x"

  • And now they seem to assert that each coset of H has the same elements. Which -if it's true- just passes my mind.

Is it that I've got the definitions wrong? That would explain why I don't understand what they are saying. Or is it that I do understand what they are saying inly that their statements are simply clear for them but not for me?

I would truly, truly appreaciate any help.

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If a coset has an element twice, then it would not have |H| elements, right? Or are they claiming that there exists no such thing as a coset with repeated elements?

A coset is first and foremost a set, so it cannot have repeated elements, in the sense that sets in mathematics are generally understood to not have repetitions, ie $\{-1,1,1\}=\{-1,1\}$. When multiplicity is taken into account, we consider multisets.

Again, I'm confused. Isn't a coset defined as $gH$ ($g$ being an element of $G$)? So, since there are $|G|$ elements we can choose from, there should be $|G$| left cosets of $H$...right? And for $|G|$ times $|H|$ to give $|G|$, $|G|$ or $|H|$ would have to be $1$. This sentence also made me become sure that I likely got some definitions wrong.

There are $|G|$ choices for $g$, but doesn't mean that different choices give rise to different cosets.

For instance, take $G=\mathbb{Z}$ the set of positive integers (with addition) and $H=2\mathbb{Z}$ the subgroup of even integers. A coset is then a set of the of the form $g+2\mathbb{Z}=\{g+2n\,|\,n\in\mathbb{Z}\}$, that is, a translation of the even integers by $g$. But it's easy to see that when $g$ is even, then $g+2\mathbb{Z}=2\mathbb{Z}$.

In fact, you can show from the definition of the equivalence relation that $g_1\sim g_2$ if and only if $g_1H=g_2H$, that is, if and only if they give rise to the same cosets.

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