I have read a few proofs and found them all really confusing, here is one for example. enter image description here

That small and simple-looking proof was really hard for me to try to understand, i have finally come to something that feels like it's what the proof tries to say, but i'm not sure it's 100% correct, I would like someone to inspect my reasonings and make sure it's correct.

(1) $a \in aH$ is perfectly understandable.

$H$ is a subgroup so it must have an identity $e \in H$.

then if we define a coset $aH = \{a \circ h: \forall h\in H\}$ we can say that $$e \in H \Rightarrow a \circ e \in aH \Rightarrow a \in aH$$

so that's good.

(2) $ aH=H \iff a \in H$ is quite more complicated.

The proof seems to have divided the $\iff$ statement into two parts, the $\Rightarrow$ statement and the $\Leftarrow $ statement, let's call them 2.1 and 2.2

2.1 for $ aH=H \Rightarrow a \in H$


we know that:

(a) $a \in aH$ as proved in (1)

(b) $aH = H \iff$ both groups have the same elements and the same operation.

if it's true that $aH = H$ then both sets have the same elements, we have proved that $a \in aH$ therefore $a \in H$. Written symbolically.

$$(aH = H) \land (a \in aH) \Rightarrow a \in H$$

2.2 for $ a \in H \Rightarrow aH=H$

To prove equality i need to prove that when the condition $a \in H$ is true it must happen that: $$ aH \subseteq H \land H \subseteq aH$$

Independently from anything related to cosets we can think this. $H$ is a subgrup, so it's closed, that means that it must hold that $$a \in H \Rightarrow a \circ h \in H :\forall h \in H $$

That implies that $a \in H \Rightarrow aH \subseteq H$ because we have arrived to the very definition of the coset $aH$, which is the one below, and every element of it is inside $H$ $$aH = a \circ h : \forall h \in H$$

now i need to prove that $H \subseteq aH$, we know that:

(a) $h \in H$ let's grab any element in $H$

(b) $a \in H$ is our condition

(c) $a \in H \Rightarrow a^{-1} \in H$ because $H$ is a group so the axiom of inverses holds

The group is closed, so $\forall h \in H $:

$$ e \circ h = H $$ $$ a \circ a^{-1} \circ h = H $$ $$ a \circ (a^{-1} \circ h) = H $$

we note that $(a^{-1} \circ h) \in H$ because of closure, let's call $$h^{2}= a^{-1} \circ h$$

we are left with the conclusion that all the elements of $H$ are in $aH$

$$ a \circ h^{2} = H $$ $$ a \circ h^{2} \in aH $$


$$H \subseteq aH$$

then by double inclusion $$ aH \subseteq H \land H \subseteq aH \Rightarrow aH = H$$

Thanks and sorry for the long post.

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    $\begingroup$ I'm not going to read all of this, but near the beginning you wrote $aH = a\circ h$. Be very careful to write thoughtful, correct sentences. The correct thing here is $aH = \{ah: h\in H\}$ (whether you include the $\circ$ or not). $\endgroup$ – Ted Shifrin Jul 6 '17 at 1:45
  • $\begingroup$ Thanks for the comment, i see the importance there, i will correct the whole thing, this difference is something that was actually causing me confusion. $\endgroup$ – Joaquin Brandan Jul 6 '17 at 1:48
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    $\begingroup$ $a\in H\implies aH= H$ is simply a consequence of $H$ being a group and groups being closed. $\endgroup$ – Doug M Jul 6 '17 at 1:54
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    $\begingroup$ You did it again, in the 4th last line of mathematics. $\endgroup$ – Theo Bendit Jul 6 '17 at 1:56
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    $\begingroup$ @JoaquinBrandan $aH$ has no repeated elements. If $ab = ac$ then $b=c.$ $\endgroup$ – Doug M Jul 6 '17 at 5:37

You're making it much more complicated than it should be.

We work under the same assumptions: $G$ is a group and $H$ is a subgroup thereof.

Lemma. For all $a\in G$, $a\in aH$.

Proof. This follows from the fact that $e\in H$ (which is a subgroup) and so, by definition, $ae\in aH$. Hence $a=ae\in aH$. QED

Proposition. If $aH=H$, then $a\in H$.

Proof. Since $a\in aH$, from $aH=H$ we deduce $a\in H$. QED

Proposition. If $a\in H$, then $aH=H$.

Proof. We need to prove two facts: $aH\subseteq H$ and $H\subseteq aH$.

First fact. Since $a\in H$, for every $h\in H$ we have $ah\in H$. Therefore $aH\subseteq H$.

Second fact. Suppose $h\in H$. We need to find $k\in H$ such that $h=ak$. Since we're in a group, we must have $k=a^{-1}h$ and this indeed belongs to $H$, because $a^{-1}\in H$ and $h\in H$. QED

Usually, the proof of the second fact doesn't mention the element $k$ you need to find, so it would just be

Suppose $h\in H$; then $h=a(a^{-1}h)$ and $a^{-1}h\in H$, so $h\in aH$.

With practice you'll get around this, in the meantime it doesn't hurt to lengthen a bit some proof (not to the level shown in your question, where you prove and reprove the same thing several times).

You can also notice that the three properties of a group have all been exploited: $e\in H$ in the lemma, closure under the operation and inverses in the second proposition.

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  • $\begingroup$ Thanks, I see that all my deductions and yours are pretty much the same, i just need no be able to explain them with the clarity and brevity you do. $\endgroup$ – Joaquin Brandan Jul 7 '17 at 17:06
  • $\begingroup$ @JoaquinBrandan Practice, practice and some more practice. ;-) $\endgroup$ – egreg Jul 7 '17 at 17:10

$aH=H \iff a \in H$.

In one direction, suppose $a \in H$. Then $ah \in H$ for all $h \in H$. Hence, $aH \subseteq H$. On the other hand, since $a \in H$, we also know that $a^{-1} \in H$. So, if $h \in H$, then $h=a\cdot a^{-1}h=a(a^{-1}h) \in aH$.

For the other direction, try the contrapositive.

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