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The book I am reading has the proves the following theorem:

Let $G$ be a group and let $H$ be a normal subgroup of $G$. The set $G/H= (aH | a \in G)$ is a group under the operation $(aH)(bH) = abH$.

The first step they show is that: $$f: G/H \times G/H \rightarrow G/H \\ aH * bH \mapsto abH$$ is well defined. I tried this verification myself, but its a little different from the one in my book. I wanted to know if it was correct:

To check if $f$ is well defined, we consider the following two mappings: $$aH * bH \mapsto abH \\ a'H * b'H \mapsto a'b'H$$ If $aH=a'H$ and $bH=b'H$ then we want $abH=ab'H$.

So suppose $aH=a'H$ and $bH=b'H$. Then $aH * bH = a'H * b'H = a'b'H$. We also know $aH * bH=abH$. Thus when $aH=a'H$ and $bH=b'H$, it follows $abH=a'b'H$.

Is this correct?

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  • $\begingroup$ yes it seems correct $\endgroup$ – Vinyl_cape_jawa Nov 28 '17 at 14:43
  • $\begingroup$ This is correct. You can write this way $abH=aH * bH=aH * bH = a'H * b'H = a'b'H$. $\endgroup$ – 1ENİGMA1 Nov 28 '17 at 14:45
  • $\begingroup$ We have $a' \in aH$ and $b' \in bH$, so $a'b' \in aHbH$. But $aHbH = a(Hb)H = a(bH)H = ab(HH) = abH$, where the second equality holds because $Hb = bH$ for any normal subgroup $H$, and the final equality holds because $HH = H$ for any subgroup $H$. We conclude that $a'b' \in abH$, or equivalently, $a'b'H = abH$. $\endgroup$ – Bungo Nov 28 '17 at 15:05
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This isn't correct. In the line "$\dots aH * bH = a'H * b'H \dots$" you explicitly assume that $f(aH,bH) = f(a'H,b'H)$, but this is what we want to prove. Another clue pointing out the mistake is that you never used the fact that $H$ is a normal subgroup of $G$, which is a must whenever $(aH)(bH) = (ab)H$ is well-defined.

To show well-definedness you have to prove that $abH = a'b'H$, without referring to the function $f$. The best way of doing this is to note that $a=a'h_1$, $b=b'h_2$. Now as $H \unlhd G$ $h_1b'=h_3b'$ and then:

$$ab = a'h_1b'h_2 = a'b'h_3h_2 = (a'b')h_3h_2 \implies ab \in (a'b')H \implies (ab)H = (a'b')H$$

or even shorter using cosets:

$$(ab)H = a(Hb) = (a'H)b = (a'b')H$$

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