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I am presented with the following problem:

Let $H = \{(), (12)(34), (13)(24), (14)(23)\}$ be a sugroup of $A_4$ (alternating group on $4$ elements), where $()$ is the identity permutation.

Find the left cosets of $H$.

My solution:

First off, I identified all the elements in $A_4$: $(1), (13)(24), (14)(23), (12)(34), (124), (132), (123), (134), (124), (143), (234), (243)$

From the multiplications below, I can conclude that the left cosets are: $[(13)(24), (124), e, (134), (234), (123), (12)(34), (142), (243), (143), (132), (14)(23)]$, which is $12$ elements. However, this contradicts with the theorem that the cardinality of the left coset is $|A_4|/|H| = 12/4 = 3$ (side note, if anybody can tell me the name of this theorem, that would be greatly appreciated). I am trouble figuring out what I did wrong, as I am pretty sure I did all the computations correctly.

Since the group operation is permutation multiplication, then we can write every element in the left coset as:

$\{ah \, \, | \,\, a \in A_4, h \in H \}$

Below are all the possible values of $ah$.

$(1) * (1, 2)(3, 4) = (1, 2)(3, 4)$

$(1) * (1, 3)(2, 4) = (1, 3)(2, 4)$

$(1) * (1, 4)(2, 3) = (1, 4)(2, 3)$

$(1, 2)(3, 4) * (1, 2)(3, 4) = (1)(2)(3)(4) = e$

$(1, 2)(3, 4) * (1, 3)(2, 4) = (1, 4)(2, 3)$

$(1, 2)(3, 4) * (1, 4)(2, 3) = (1, 3)(2, 4)$

$(1, 3)(2, 4) * (1, 2)(3, 4) = (1, 4)(2, 3)$

$(1, 3)(2, 4) * (1, 3)(2, 4) = (1)(2)(3)(4) = e$

$(1, 3)(2, 4) * (1, 4)(2, 3) = (1, 2)(3, 4)$

$(1, 4)(2, 3) * (1, 2)(3, 4) = (1, 3)(2, 4)$

$(1, 4)(2, 3) * (1, 3)(2, 4) = (1, 2)(3, 4)$

$(1, 4)(2, 3) * (1, 4)(2, 3) = (1)(2)(3)(4) = e$

$(1, 2, 3) * (1, 2)(3, 4) = (1, 3, 4)$

$(1, 2, 3) * (1, 3)(2, 4) = (1)(2, 4, 3) = (2, 4, 3)$

$(1, 2, 3) * (1, 4)(2, 3) = (1, 4, 2)$

$(1, 2, 4) * (1, 2)(3, 4) = (1, 4, 3)$

$(1, 2, 4) * (1, 3)(2, 4) = (1, 3, 2)$

$(1, 2, 4) * (1, 4)(2, 3) = (2, 3, 4)$

$(1, 3, 2) * (1, 2)(3, 4) = (2, 3, 4)$

$(1, 3, 2) * (1, 3)(2, 4) = (1, 2, 4)$

$(1, 3, 2) * (1, 4)(2, 3) = (1, 4, 3)$

$(1, 3, 4) * (1, 2)(3, 4) = (1, 2, 3)$

$(1, 3, 4) * (1, 3)(2, 4) = (1, 4, 2)$

$(1, 3, 4) * (1, 4)(2, 3) = (2, 4, 3)$

$(1, 4, 2) * (1, 2)(3, 4) = (2, 4, 3)$

$(1, 4, 2) * (1, 3)(2, 4) = (1, 3, 4)$

$(1, 4, 2) * (1, 4)(2, 3) = (1, 2, 3)$

$(1, 4, 3) * (1, 2)(3, 4) = (1, 2, 4)$

$(1, 4, 3) * (1, 3)(2, 4) = (2, 3, 4)$

$(1, 4, 3) * (1, 4)(2, 3) = (1, 3, 2)$

$(2, 3, 4) * (1, 2)(3, 4) = (1, 3, 2)$

$(2, 3, 4) * (1, 3)(2, 4) = (1, 4, 3)$

$(2, 3, 4) * (1, 4)(2, 3) = (1, 2, 4)$

$(2, 4, 3) * (1, 2)(3, 4) = (1, 4, 2)$

$(2, 4, 3) * (1, 3)(2, 4) = (1, 2, 3)$

$(2, 4, 3) * (1, 4)(2, 3) = (1, 3, 4)$

Justin Trudeau
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    A coset is a subset of the (big) group, not an element of it. One coset is the subgroup itself. – Arthur Oct 18 '21 at 09:24

1 Answers1

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As @Arthur points out, cosets are sets not elements. In fact in your own question you have the answer:

The cosets associated to $()$, $(12)(34)$, $(13)(24)$ and $(14)(23)$ are $H$.

Those corresponding to $(2,4,3)$, $(1,3,4)$, $(1,2,3)$ and $(1,4,2)$ are, well, they form the coset.

The rest is the other coset $\{(1,2,4), (1,3,2), (1,4,3), (2,3,4)\}$.

You can check with in your calculations how $(1,2,4) H = (1,3,2) H$ for example.

Moreover, since cosets form a partition of the large group $A_5$ an element $x$ gives the same coset as another $y$ if and only if $x\in yH$. Therefore, you know $H$ is a coset. Now, take any point $x\in A_5\smallsetminus H$ and compute its coset $xH$. Now, take another $y\in A_5\smallsetminus (H\cup xH)$ and do the same until you have covered all $A_5$.

It is called Lagrange's Theorem.

Hope it helps.

SrRojizo
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    Thanks @SrRojizo! I was just wondering, is there a more efficient to compute the cosets then going through all the combination of possible values for $a$ and $h$? – Justin Trudeau Oct 18 '21 at 10:20
  • @justin, you could same some steps as I point out in the answer. $H$ is a coset. Then take $x\in G$ but $x\notin H$. Then $xH$ is another (different) coset. Now take another $y\in G$ but $y\notin H\cup xH$ and continue in this way... I do not know any smarter way... By the way it may be better explained in this answer – SrRojizo Oct 19 '21 at 19:51