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I am reading DF's proof of Lagrange's theorem that the order of a subgroup divides the order of a group.

The part where I cannot follow is when they say that :

The set of left cosets of H in G partition G (I can see this).

Bythe definition of a left coset the map:

$H \rightarrow gH$ defined by $h \mapsto gh$ is a surjection from H to the left coset gH (this I cannot see clearly) but I know the definition of a left coset is that you take all elements of G and multiply on the subgroup H on the left side

Injective since $gh_{1}=gh_{2} \implies h_{1}=h_{2}$

Then they conclude that H and gH have the same order I guess this is because we can see now it is bijective?

My question here is about the Surjectiveness, how can I see that? and my other question is what excatly are they doing when showing injectiveness are they just looking on the element map? \mapsto

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    $\begingroup$ Surjectiveness is by definition! After all, $gH =\{gh \mid h\in H\}$. As for the injectiveness, that's how you prove it in general: Assume two inputs map to the same thing and show they are equal. $\endgroup$
    – John Brevik
    Commented Apr 4, 2018 at 10:32
  • $\begingroup$ so by definition for all elements of the left cosets there exists elements in the subgroup H s.t. $H \rightarrow gH$ . I think what confuse me is it seems like they says that the surjectiveness is shown in the group mapping, but the injectiveness is shown in the element mapping $\endgroup$
    – user420309
    Commented Apr 4, 2018 at 10:40
  • $\begingroup$ $H \to gH$ is not "the group mapping". In the proof, it is D+H's way of denoting that the domain of the coset map is $H$, and the codomain of the coset map is $gH$. $\endgroup$
    – Ben Grossmann
    Commented Apr 4, 2018 at 10:46

2 Answers 2

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Having looked at your post and flipped to the proof from Dummit and Foote that you're talking about, I have a hunch that you're getting hung up on the notation. That being said, perhaps it will help to rephrase the major points.

For any element $g \in G$, we can define the left-coset map $$ \phi_g : H \to gH\\ \phi_g(h) = gh $$ The image of this map is $$ \phi_g(H) = \{\phi_g(h) : h \in H\} = \{gh: h \in H\} $$ so that, by the definition of a left coset, $\phi_g(H) = gH$. In other words, the map $\phi_g$ is surjective.

Now, try to use the left-cancellation law to show that $\phi_g$ is injective.

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  • $\begingroup$ Thanks so in their proof $h_{1}$ and $h_{2} \in H$ ? In other words it doesn't matter which element you pick from H if you multiply by $g \in G$ $gh_{1}=gh_{2} \implies h_{1}=h_{2}$? $\endgroup$
    – user420309
    Commented Apr 4, 2018 at 12:59
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    $\begingroup$ @Djhoe that's right $\endgroup$
    – Ben Grossmann
    Commented Apr 4, 2018 at 13:01
  • $\begingroup$ Do you mind if I ask you again,... Why do they care to show that H and gH have the same cardinality.. Is it because they have defined the number of left cosets of H in G to equal K and then they end up saying k=|G|/|H| $\endgroup$
    – user420309
    Commented Apr 4, 2018 at 13:21
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    $\begingroup$ @Djhoe yes, that's exactly right $\endgroup$
    – Ben Grossmann
    Commented Apr 4, 2018 at 14:00
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I think misunderstanding is in statement "the definition of a left coset is that you take all elements of G".

No, definition if coset is that you take one element of G and multiply it by all element of $G$ (although different elements $g_1$ and $g_2$ can generate same coset). By taking all element of $G$ you create all cosets and $H$ itself (if $g \in H$).

So, function $H \rightarrow gH$ is surjective because for every $h$ exist element in $gH$ (it is $gh$), and as was mentioned by you injective

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