Question: Multiplication rule for normal subgroup $H\triangleleft G$ $$aH\circ bH=(ab)H$$ is well-defined?

Proof(1): (From lecture notes) Suppose $a'H=aH,b'H=bH$ need to check $a'b'H=abH$. So $a'=ah_1,b'=bh_2,a'b'=ah_1bh_2=ab(b^{-1}h_1b)h_2)\in abH$

Proof(2): (From youtube video https://www.youtube.com/watch?v=lVYV_34mL6Y&list=PLAvgI3H-gclb_Xy7eTIXkkKt3KlV6gk9_&index=41) Suppose $x'\in aH,y'\in bH$ need to check $x'y'\in abH$. So $x'=ah_1,y'=bh_2,x'y'=ah_1bh_2=ab(b^{-1}h_1b)h_2)\in abH$

Proof(3): (My own) $x'y'=ah_1bh_2=abh_1'h_2\in abH$ since $aH=Ha$ by normality there exist $h_1'$ that commutes

Are all three proofs above correct? What's the difference between the proof (1) and proof (2)? (proof (1) doesn't seem legit to me at $a'H=aH\implies a'=ah_1$) I know there are questions here and wikipedia regarding how to proof well-definedness in group theory but if proof (2) and proof (3) are correct then we're not really choosing two different representatives? I think for different representatives to be chosen from $aH$ they should at least share the same $a$ instead of different $a$ but $h$ might be chosen differently so I think the right way is to choose $ah_1,ah_2\in aH$ and $bh_1,bh_2\in aH$ and check if $ah_1bh_2,ah_3,bh_4$ lies in the same $(ab)H$?

There are similar questions like In Group theory proofs what is meant by "well defined" but I don't feel like they answered the questions I'm wondering so I choose to post this.

  • 3
    $\begingroup$ There is absolutely no difference between the three proofs, which are all correct. It's a good exercise to convince yourself that the three proofs are saying the exact same thing but with different words, you should try it before reading the answers to your question. $\endgroup$ Nov 6, 2016 at 21:02
  • $\begingroup$ @NajibIdrissi can you answer why $a'H=aH\implies a'=ah_1$ in the first proof? $\endgroup$
    – ZHU
    Nov 6, 2016 at 21:05
  • $\begingroup$ @ZHU If $a' H = aH$, then for any $h'$ there is some $h''$ such that $a'h' = ah''$. Multiplying on the right by $(h')^{-1}$ and renaming $h'' (h')^{-1} = h_1$ gives that $a' = ah_1$. $\endgroup$
    – davidlowryduda
    Nov 6, 2016 at 21:10
  • $\begingroup$ Note that $a'$ is an element of $a'H$, since $H$ contains the identity element. Therefore, if $a'H = aH$ then $a' \in aH$, which implies that $a' = ah_1$ for some $h_1 \in H$. $\endgroup$ Nov 6, 2016 at 21:10
  • $\begingroup$ @mixedmath my concern is that every left coset is given by multiplying a unique element $a$ to $H$ if $a\not\in H$. But how can they be possibly chosen differently from one another? $\endgroup$
    – ZHU
    Nov 6, 2016 at 21:17

1 Answer 1


I write this comment as an answer for lack of space. The calculation you mention is in fact more symbolic than real. In reality one multiplies whole sets (the cosets) with each other in the sense that $aH$ stands for the whole set $\{ah_1, ah_2, \ldots \} = \{ah|h \in H\}$. Multiplying two cosets $aH$ and $bH$ then consists in making a new set consisting of all the possible products $ah_ibh_j$. Now this makes no sense if $H$ is not normal because this set is no coset anymore but if $H$ is normal we end up with the set $\{abh'_ih_j|h'_i,h_j \in H\} = \{abh|h \in H\}$. What one does in those proofs is: one picks one representative in each coset, multiplies them and obtain a representative of a new coset. What is not obvious in these proofs is that one obains the same coset if one had chosen other representatives. A good way to visualize this situation is to take $G = \Bbb{R}^2,_+$ the group of vectors in the plane, and $H$ a 1-dimensional subspace (line through the origin). Then the cosets are lines parallel to this line. If one picks a point on each line and adds up the vectors defined by these points one obtains a vector pointing into always the same line, independent from the chosen points.


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