I am working on through a proof from lecture. Given that $K\leq H\leq G$, we are constructing a bijection $$f: G/H\times H/K\rightarrow G/K$$ Since $G/H$ is the set of cosets, we can fix the representative of the cosets in creating a well defined function. That is $$g_1H = g_2H$$ if and only if $g_1 = g_2$.

That last part about fixing the representatives of each coset leaves me uneasy. Since we know that multiple representatives prevents us from creating an operation on any set of cosets, I am not sure why we can fix our representatives.

If that is true then the rest of the proof follows.Why are we allowed fix representatives of the cosets for this function. Is it related to mapping cosets from one set to a different set?

  • $\begingroup$ If you pick a representative and use it to define a function, then you need to show that if you picked a different representative from the same class, then you would get the same output to your function. Is this your worry? $\endgroup$
    – James
    Oct 13 '18 at 0:11
  • $\begingroup$ @James My professor said that we could fix our representatives for the cosets so that a set is only represented by one element and we would then not truly need to check if a coset was represented by a different element. Which I think is wrong, or at least is very dangerous and I do not know why we can use it in this case. $\endgroup$ Oct 13 '18 at 0:15
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    $\begingroup$ I don't think I agree with @James here. Choosing different representatives will give you a different bijection. But that's okay; you're only asked to find one bijection. $\endgroup$
    – Billy
    Oct 13 '18 at 0:21
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    $\begingroup$ Yes. The point is that $f$ depends on those representatives. If $R$ and $R'$ are two sets of representatives for the cosets, then the functions $f_R$ and $f_{R'}$ you get out of them will usually be different. Both will be bijections, they just won't be the same. This would only be a well-definedness issue if we were trying to claim that $f_R = f_{R'}$. $\endgroup$
    – Billy
    Oct 13 '18 at 0:26
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    $\begingroup$ It's not the job of the function to "identify" anything - it's our job to give the function an exhaustive and unambiguous list of rules. (For instance, "if you are given coset $C_i$, you should proceed using representative $x_i$".) Perhaps an example will make this clearer, so I'll post one in an answer. $\endgroup$
    – Billy
    Oct 13 '18 at 0:34

As I said in the comments above: $f$ will depend on the choice of the set of representatives.

I think the best way to see what's going on is with an example, so let

  • $G = \{e, g, g^2, g^3, g^4, g^5\}$, a cyclic group of order 6,
  • $H = \{e, g^3\}$, the unique subgroup of order 2,
  • $K = \{e\}$.

$G/H$ is a cyclic group of order 3, and we could pick a bunch of different lists of our favourite representatives. Two possible choices are, for example,

  • $R^{G/H}_1 = \{e, g, g^2\}$
  • $R^{G/H}_2 = \{e, g, g^5\}$

-- that is, I've just swapped the $g^2$ for a $g^5$.

$H/K$ is a cyclic group of order 2, and only has one possible set of representatives: $\{e, g^3\}$. (This is because I've taken $K$ to be trivial, for the sake of simplicity, so that we can just ignore it in this example.) So we're forced to set both $R^{H/K}_1$ and $R^{H/K}_2$ -- our lists of our chosen representatives of $H/K$ -- equal to this set.

We can now define our functions

$$f_1, f_2: G/H \times H/K \to G/K$$

as follows: given some cosets $xH\in G/H$ and $yK\in H/K$, the function $f_1$ will look up our chosen representative for $xH$ in the list $R^{G/H}_1$, and our chosen representative for $yK$ in $R^{H/K}_1$; and $f_2$ will do the same, but using the lists $R^*_2$. Both functions then proceed to multiply these elements in order and stick a $K$ at the end and output the result.

So, for instance: consider the cosets $(eH, eK)$. $f_1$ will look up the representatives in lists $R^*_1$, and find that they're $e$ and $e$; $f_2$ will look up the representatives in $R^*_2$ and get the same answers. The result will be $f_1(eH, eK) = f_2(eH, eK) = e\cdot e\cdot K = eK$.

But consider the cosets $(g^2H, eK)$. When $f_1$ looks up the representatives, it finds $g^2$ and $e$; but when $f_2$ looks them up, it finds $g^5$ and $e$. So $f_1(g^2H, eK) = g^2K$, but $f_2(g^2H, eK) = g^5K$. These aren't equal (as elements of $G/K$).

You can visualise the outputs of $f_1$ and $f_2$ by drawing a table for each: 3 rows (one for each coset of $G/H$), two columns (for cosets of $H/K$). You'll notice that, in both cases, all 6 outputs are distinct - in other words, both are injective.

  • $\begingroup$ So this creates many functions for the specific case that we are looking at and each one of these functions is a bijection. This is the part where I am getting confused, how does having bijections for every possible case show that we then have a bijection from $G/H\times H/K \rightarrow G/K$ $\endgroup$ Oct 13 '18 at 1:06
  • $\begingroup$ I don't understand your question. $f_1$ is a bijection $G/H\times H/K\to G/K$. So is $f_2$. Any one of them will do. $\endgroup$
    – Billy
    Oct 13 '18 at 1:14
  • $\begingroup$ I guess, why is ok that our function is dependent on the input for its definition. Why do we not need a function that works for every input. $\endgroup$ Oct 13 '18 at 1:17
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    $\begingroup$ I think its starting to click, to give a different example. If we made a function $g: \mathbb{Q}\rightarrow \mathbb{N}$ and defined $g(p/q) = p + q$, but we require that $p$ and $q$ are relatively prime. $\endgroup$ Oct 13 '18 at 1:29
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    $\begingroup$ Yes, that's exactly the sort of thing that's going on. $\endgroup$
    – Billy
    Oct 13 '18 at 1:30

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