3

In Fraleigh, there is what appears to be a classic theorem on cosets. I'm confused about the proof for the converse.

Theorem. Let $H$ be a subgroup of a group $G$. Then left coset multiplication is well defined by the equation $$(aH)(bH) = (ab)H$$ if and only if $H$ is a normal subgroup of $G$.

Proof. ...We now turn to the converse: if $H$ is a normal subgroup, then left multiplication by representatives is well-defined. Suppose we wish to compute $(aH)(bH)$. Choosing $a \in aH$ and $b \in bH$, we obtain the coset $(ab)H$. Choosing different representatives $ah_1 \in aH$ and $bh_2 \in H$, we obtain the coset $ah_1 b h_2 H$. We must show these are the same coset. Now $h_1 b \in H b = bH$, so $h_1b = bh_3$ for some $h_3 \in H$. Thus $$(ah_1)(bh_2) = a(h_1 b) h_2 = a(bh_3)h_2 = (ab)(h_3h_2)$$ and $(ab)(h_3h_2) \in (ab)H.$ Therefore $ah_1 b h_2$ is in $(ab)H$.

This seems weird to me. Why do we name our second representatives $ah_1$ and $bh_2$, rather than $a', b' \in G$? For example, in this post Arturo produces a proof for this,

Proof. Suppose $H$ is normal. Then $gH=Hg$ for every $g\in G$. If $xH=x'H$ and $yH=y'H$, then $$xyH = x(yH) = x(Hy) = x(Hy') = (xH)y' = > (x'H)y' = x'(Hy') = x'(y'H) = x'y'H$$ so the operation is well defined.

There must be some motivation for using this seemingly awkward way of picking representatives.

Community
  • 1
xzackli
  • 288

2 Answers2

3

This difference probably just arises as a result of different expositional choices by the authors about "cosets," but at any rate both explanations work.

If Fraleigh defines $aH=\{ag\mid g\in H\}$, then his description makes sense, because that's the only description you have of $aH$.

Arturo on the other hand might choose to present $aH$ to be the equivalence class of things under the relation $x\sim y\iff x^{-1}y\in H$. In that case, when Arturo writes $aH=bH$, we implicitly have $b=a(a^{-1}b)=ah$.

While Arturo's expression is more compact, Fraleigh's notation does have one benefit. Some beginners mistakenly believe that $aH=Ha$ means that $ah=ha$ for all $h\in H$. By going down to the level of $h_i$'s in his notation, Fraleigh is clear that no such commutation is taking place:

$$(ah_1)(bh_2) = a\underline{(h_1 b)} h_2 = a\underline{(bh_3)}h_2 = (ab)(h_3h_2)$$

With the "simplified" notation, beginners might overlook that detail.

rschwieb
  • 153,510
2

Remember, too, that $aH =\{ah:h\in H\}$. So his first choice of representatives weren't just $a$ and $b$, they really were of the form $ah'_1$ and $bh'_2$ (take $h'_1 = h'_2 =e$) like the second choice of representatives $ah_1$ and $bh_2$.

MPW
  • 43,638