# Is this a valid proof of Lagrange's theorem (finite case).

Let $$G$$ be finite and $$H$$ be a subgroup. We will show that the left cosets of $$H$$ partition $$G$$ and each coset has the same size.

1) Each element $$g \in G$$ belongs to the coset $$gH$$ since $$g1=g$$ and $$1\in H$$.

So every element lies in at least one coset.

2) We know show that each $$g \in G$$ lies in exactly one coset.

Suppose for a contradiction $$g$$ lies in more than one coset then $$g \in aH$$ and $$g \in bH$$ where $$aH,bH$$ are distinct left cosets. Then $$g=ah_1$$ and $$g=bh_2$$ so $$ah_1=bh_2$$ for some $$h_1,h_2 \in H$$.

Now $$aH \subseteq bH$$ since if $$ah \in aH$$ then $$ah=bh_2h_1^{-1}h \in bH$$. Likewise $$bH \subseteq aH$$ so $$aH=bH$$ a contradiction.

So each $$g \in G$$ lies in exactly one coset.

Hence the cosets form a partition of $$G$$.

3) For each $$g \in G$$ the coset $$gH$$ has the same order as $$H$$.

To see this establish a function $$\phi:gH \rightarrow H$$ by $$\phi(gh)=h$$. This is clearly a bijection.

So $$|G|=\text{Number of cosets} \times \text{Size of each coset}=[G:H]|H|$$ and so $$|H| \mid |G|$$.

Is this proof valid?

• What do you mean "finite case"? Lagrange's theorem is only applicable to finite groups, since "divides the order" only makes sense when "order" is a number. – Adam Hughes May 26 '17 at 14:11
• There's just a small issue with your argument that $aH\subseteq bH$. You wrote $ah = bh_2h_{1}^{-1}$, where I think it should be $ah = bh_2h_{1}^{-1}h$. – James May 26 '17 at 14:22
• By finite case I meant there is some kind of extension that the index is infinite for infinite groups. – Ben B May 26 '17 at 15:36
• @James Yes you are correct that was just a mistake when writing it out. Is everything else sound though? Thanks! – Ben B May 26 '17 at 15:37
• Everything sounds good to me. Though it may be easier to justify the last point by considering $\phi : H\to gH$ defined by $\phi(h)= gh$. It would then be easier to justify that :1. It is well-defined, 2. It is bijective – Maxime Ramzi May 26 '17 at 17:08

Your proof is fine, except that I recommend you consider \begin{align} \varphi: H&\to gH \\ h&\mapsto gh, \end{align}
then justify that $$\varphi$$ is a well-defined bijection; it's much easier than your $$\phi$$.