I understand that H= {e, (123), (132)} and ord(H)=3. And S4 has 24 elements since 4!=24 so 24/3 means there are going to be 8 distinct cosets. I'm stuck on the multiplying part and if you say let g=(1234), then multiply gH for the first coset, then g^2H for the second coset? I'm confused as to how to find all 8 cosets.
[Math] Compute the left cosets of <(123)> in (S4, o).
abstract-algebragroup-theorysymmetric-groups
Related Solutions
"What are the left and right cosets of $D_4$?" does not make sense. You have to ask "what are the left and right cosets of H in G?", where H is some subgroup of G. NasuSama's answer has given you a good way of calculating the cosets of $\{e, y^2\}$ in $D_4$, but I suspect you wanted to know the cosets of $D_4$ in $S_4$.
First of all, cycle notation. You should think about $(1234)$ as a function, let's call it $f$, such that
- $f(1) = 2$
- $f(2) = 3$
- $f(3) = 4$
- $f(4) = 1.$
Similarly, $(13)(24)$ is a function, let's call it $g$, such that
- $g(1) = 3$
- $g(2) = 4$
- $g(3) = 1$
- $g(4) = 2.$
Do you see why? Now, when you "multiply" f by g, you simply find the function $fg$ (or $f\circ g$ if you prefer). So $fg(1) = f(g(1)) = f(3) = 4$, and so on. (Warning: some people write $fg$ to mean $gf$. Both notations are in use - check with your lecturer or textbook.)
Do you now understand how to form cosets? To find the left coset of $D_4$ in $S_4$ corresponding to the element $(123)$, just left-multiply everything in $D_4$ by $(123)$.
Here are a few helpful facts about cosets of $H$ in $G$:
- Any two left cosets are either exactly the same, or completely disjoint.
- If $h\in H$, then $hH = H$.
- If $g\in G$ but $g\not\in H$, then $gH \neq H$.
- If $g_2 \in g_1H$, then $g_1H = g_2H$.
Everything I said above works with left cosets replaced by right cosets. Be careful when you mix left with right, though:
- It is not generally true that $g_1H$ and $Hg_2$ must be disjoint or equal.
- It is not generally true that $g_1H = Hg_1$.
The permutation $(1)$ simply means the identity, which sends:
$1 \mapsto 1\\2 \mapsto 2\\3 \mapsto 3.$
We could also denote it by $(1)(2)(3)$ (three disjoint $1$-cycles), which might be a bit less confusing. It's common practice to omit $1$-cycles from permutations, but in the case of the identity map, that leaves us "nothing to write down" (which of course is what the identity map does-namely, "nothing").
If you apply the permutation $(1\ 2)$ to each element of $H$, which is what is meant by:
$(1\ 2)H$, you obtain (applying the pemutation on the right first, as is often done with composition):
$(1\ 2)(1) = (1\ 2)$
$(1\ 2)(1\ 2\ 3) = (2\ 3)$
$(1\ 2)(1\ 3\ 2) = (1\ 3)$, which is what you surmised.
It's not that hard to see that $\{(1), (1\ 2\ 3), (1\ 3\ 2)\}$ form a cyclic subgroup of order $3$, which we might also denote as:
$H = \{1,a,a^2\}$ (since $(1\ 2\ 3)(1\ 2\ 3) = (1\ 3\ 2)$).
then $(1)H = 1H = H$ (this is obvious), while:
$(1\ 2\ 3)H = aH = \{a,a^2,a^3 = 1\}$, which is the same SET (the order of elements doesn't matter in a set) as $H$.
Similarly, $(1\ 3\ 2)H = a^2H = \{a^2,a^3 = 1,a^4 = a^3a = a\} = H$.
It turns out in this case (and you might well speculate as to what happens in general) that for $\sigma \in S_3,$ we have:
$\sigma H = H \iff \sigma \in H$
Best Answer
Finding the cosets in this small case is not so bad. First and foremost you have the coset $$H=\{e,(123),(132)\}$$ Here is a general algorithm for how to finish. So far, say you've computed $k<8$ cosets. Pick an element of $S_4$ that is not in any of the cosets you've computed so far, and that will give you a new coset.
With this method I'll find one more coset for you. The cosets we have are $$\{e,(123),(132)\}$$ $$\{(1234),(1234)(123)=(2413),(1234)(132)=(14)\}$$ None of the two cosets we've found contain $(234)$. So we multiply the elements of $H$ on the left by $(234)$ to get a new left coset, $$\{(234),(24)(13),(142)\}$$ Every element of the group is in exactly one coset, so continuing in this way will get you to the end.