working on a practice question about finite abelian groups and just want to see if I am on the right track:
Let $H = <(123)(4567),(8\space 9)(10\space 11),(8\space 11)(9\space 10) > \space \le
\sum_{11}$(a) Show that $H$ is abelian and determine the order.
(b) Give a complete list (without duplications) of all isomorphism
classes of abelian groups having order $|H|$.(c) Determine the isomorphism class of $H$.
For (a), To show that $H$ is abelian, I simply show that each element in the generator of $H$ is commutative with one another. Since every element in $H$ is a combination of these elements – this should be sufficient to show $H$ is abelian.
To determine the order of $H$ I again consider each element of the generator of $H$. If I let $a= (123)(4567)$, $b= (8\space 9)(10\space 11)$, $c=(8\space 11)(9\space 10)$.
Then $a$ has order 12, and $b$, $c$ have order 2. Thus the order of $H$ is 48. With the elements: $a^i$,$a^ib$, $a^ic$, $a^ibc$ ($i = 0,1,…,11$). Have I got the right idea here?
For (b):
Since the prime factorisation of $48 = 3 \cdot 2^4$, the complete list is:
$C_3 \times \space C_2 \times \space C_2 \times \space C_2 \times \space C_2 $
$C_3 \times \space C_2 \times \space C_2 \times \space C_{2^2}$
$C_3 \times \space C_2 \times \space C_{2^3}$
$C_3 \times \space C_{2^4}$
$C_3 \times \space C_{2^2} \times \space C_{2^2}$
For (c):
I'm not too sure about this, but does it not just follow from theorem that from the prime factorisation that $H \cong C_3 \times \space C_{2^4}$? (Indeed I did find subgroups of order $3$ and $2^4$, whose direct product is $H$) But do you need to prove that $H$ is not isomorphic to the other classes?
Many thanks in advance.
Best Answer
Parts (a) and (b) look fine. Though, if this were graded work, I'd want to see a little more argument as to why your enumeration of the elements of $H$ (to determine the order) was complete and non-redundant.
There seems to be a problem with (c). It's fairly clear that $H$ is the direct product $\langle a\rangle\times\langle b,c\rangle$, since the permutation $a$ acts on a different set of points that do $b$ and $c$. You seem to have already realised that $\langle b,c\rangle\cong C_2\times C_2$. So your group is $C_{12}\times C_2\times C_2$. But the $2$-part of the $C_{12}$ yields a further direct factor ($C_4$) for the $2$-part of the whole group $H$. So the end result should be $C_3\times C_4\times C_2\times C_2$. (I.e., the second on your list.) It should not be necessary to argue that it is not isomorphic to any of the others. Your list represents a complete and non-redundant enumeration (up to isomorphism) of the abelian groups of order $48$.