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If $G$ is a group and $H$ a subgroup of $G$, then the set $gH = \{gh \mid h \in H\} = \{g, gh_1, gh_2,\ldots\}$ is a "left coset" of $G$ wrt $H.$

So does this basically mean, that when I multiply some element of group $H$ on the left by some element from the group $G$, this is a left coset? Doing this multiple times with multiple elements from $G$ and $H$ will give me the set of left cosets? The same for right cosets.

Why is this important then? If the cosets have a one to one relation (like in the Orbit Stabiliser theorem), does that imply a bijection? As one element from the group $G$ is mapped to only another element in $H$ and each of these "outcomes" are unique?

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    $\begingroup$ When you write "Doing this multiple times with multiple elements from G and H..." indicates your confusion. $gH$, is all elements of $H$ multiplied by one element of $g$ on the left. "As one element from the group G is mapped to only another element in H and each of these "outcomes" are unique," again, I have no idea what you mean. It would help probably to look at examples in the simplest group, $\mathbb Z$. What are the cosets of the subgroup of even integers? $\endgroup$
    – Thomas Andrews
    Commented Dec 30, 2012 at 17:48
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    $\begingroup$ You can think as $H$ being 'rotated' by the element $g$. Each element $g$ 'rotates' the whole subgroup $H$ by multiplication. $\endgroup$
    – Sigur
    Commented Dec 30, 2012 at 17:53
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    $\begingroup$ Create some examples. Find all the left and right cosets of $\mathbb{Z}_7$ and $D_4$, or any two small, finite groups, one Abelian and one not. See what you get, see what's different, etc. If that makes sense then you might play around with a group that has all normal subgroups and make a quotient group. Create whatever tables or diagrams you need to see what's going on. It might take a while but it is through that kind of work that the most solid understand is acheived. $\endgroup$
    – Todd Wilcox
    Commented Dec 30, 2012 at 18:06

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If you "multiply some element of $H$ on the left by some element from the group $G$", that is not a coset. If you multiply all elements of $H$ on the left by one element of $G$, the set of products is a coset.

If $H$ happens to be a normal subgroup (i.e. its left cosets are the same as its right cosets), then one can actually multiply cosets, and that gives another group, the quotient group $G/H$.

(I'm having trouble figuring out what you're trying to say in your last paragraph.)

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    $\begingroup$ My last part is just "whats so important about them?". Right now I'm going through proving OS theorem and I need to prove that the Orbit has 1 - 1 relation with cosets of G, but what does this mean? Why does proving that help me prove the OS theorem? I use a similar thing to prove Lagrange theorem. How do these cosets help with these proofs? $\endgroup$
    – Kaish
    Commented Dec 30, 2012 at 19:15
  • $\begingroup$ A group $G$ acts on a set $X$. Let $x$ be one element of $X$. Let $H$ be the set of all $h\in G$ such that $hx=x$. Then $H$ must be a subgroup. To see that, suppose $h_1,h_2\in H$. Then $(h_1h_2)x=h_1(h_2x)$ (this follows from the definition of group action) $=h_1x = x$. Therefore $h_1h_2\in H$. Now suppose $h\in H$ and think about $h^{-1}$. We have $x=(h^{-1}h)x=h^{-1}(hx)=h^{-1}x$. Therefore $h^{-1}\in H$. So how do we see that for any two $g_1,g_2\in G$, we have $g_1x=g_2x$ if and only if $g_1,g_2$ are both in the same coset of $H$? A similar argument does it. $\endgroup$
    – Michael Hardy
    Commented Dec 30, 2012 at 19:25
  • $\begingroup$ . . . . and that's one occasion for thinking about the concept of "coset". $\endgroup$
    – Michael Hardy
    Commented Dec 30, 2012 at 19:25
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A left coset is an equivalence class of $G/\sim$, where $\sim$ is the equivalence relation that states that two elements of the group, $g_1$ and $g_2$, are equivalent if $g_1 = g_2 h$ for some element $h \in H$. This is equivalent to your definition as the set $\{gH : g \in G\}$, since if $g_1 \sim g_2$ we have $g_1 H = g_2 H$, and vice versa. If you can understand why it might be useful to understand equivalence classes, it will become clear why it is interesting to study cosets.

To be more concrete with the computations, suppose that $g_1 \sim g_2$, so that $g_1 = g_2 h$. Then we have $g_1 H = g_2 h H$. Thus, it will suffice to show that $hH = H$. Since $H$ is a subgroup, can you see why multiplying every element on the left by $h \in H$ only shuffles around all the elements, without losing anything? Conversely, if $g_1 H = g_2 H$ as sets, then we know that the left hand side contains an element of the form $g_1 e = g_1$, which must be equal to some element $g_2 h$ on the right hand side.

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Let's go back to basic set theory. How many elements are in the set:

$$S=\{\{1,2\},\{3,4\}\}$$

If you said $4$, you'd be wrong, there are two - the elements are sets, $\{1,2\}$ and $\{3,4\}$.

While a coset is a set, when we talk about "the set of all cosets," we are actually talking about a set containing sets as members.

You keep talking about individual members of $H$. That is essentially useless. $H$ is a "block" that is acted on uniformly by a single $g$, not individually having different $h\in H$ acted on by different $g\in G$.

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