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?

  • 1
    $\begingroup$ 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. $\endgroup$ – Adam Hughes May 26 '17 at 14:11
  • $\begingroup$ 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 $ah = bh_2h_{1}^{-1}h$. $\endgroup$ – James May 26 '17 at 14:22
  • $\begingroup$ By finite case I meant there is some kind of extension that the index is infinite for infinite groups. $\endgroup$ – Ben B May 26 '17 at 15:36
  • $\begingroup$ @James Yes you are correct that was just a mistake when writing it out. Is everything else sound though? Thanks! $\endgroup$ – Ben B May 26 '17 at 15:37
  • $\begingroup$ 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 $\endgroup$ – Maxime Ramzi May 26 '17 at 17:08

To summarise the comments . . .

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 $\phi$.

Also, Lagrange's Theorem only applies to finite groups.

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