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Exercise: Let $H$ be a subgroup of the finite group $G$ and let $H$ act on $G$ (here $A = G$) by left multiplication. Let $x \in G$ and let $O$ be the orbit of $ x$ under the action $H$.

Prove that the map $H → O$ defined by $h → hx$

is a bijection (hence all orbits have cardinality $|H|$)

Hint: if $H$ is a group acting on a set $A$, the relation ~ on $A$ defined by $a$~$b$ if and only if $a = bh$ for some $h \in H$ is an equivalence relation.

Then using the hint, and the bijection deduce Lagrange's Theorem: "If $G$ is a finite group and $H$ is a subgroup of $G$ then $|H|$ divides $|G|$.

attempt: We will show map $H → O$ defined by $h → hx$

is a bijection . Hence we need to show it's injective and surjective. For injectivity , suppose $h_1, h_2 \in H$ , then we need to show $\phi(h_1) = \phi (h_2) → h_1 = h_2$. Then $h_1x = h_2x$ implies $(h_1x)x^{-1} = (h_2x)x^{-1}$ implies $h_1(xx^{-1}) = h_2(xx^{-1})$ implies $h_1 = h_2$ Thus it's injective.

for surjective, clearly it's onto since for every $hx \in O $, there is an $h \in H$ such that $\phi(h) = hx$. Therefore it's a bijection.

Can someone please help me deduce Lagrange's Theorem, using the fact that $\phi$ is a bijection and using the given hint? Thank you very much.

Community
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Mahidevran
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    That's the approach suggested in my pet question https://math.stackexchange.com/questions/28332/is-lagranges-theorem-the-most-basic-result-in-finite-group-theory. – lhf Sep 10 '15 at 02:52

1 Answers1

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By your bijection, it follows that $$ |O_x| = |H| \quad\forall x\in G $$ where $O_x$ is the orbit of $x$. Now note that $O_x = Hx$ (the right coset) and that $G$ is partitioned into orbits. Hence, $$ |G| = \sum_{x\in A} |O_x| $$ for some subset $A\subset G$. In particular, $|G| = |A||H|$, so $|H| \mid |G|$.

Prahlad Vaidyanathan
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