1
$\begingroup$

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?

$\endgroup$
  • 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
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.