0

Seems similar to this question, but I want to calculate it for $n = 1\ldots6$. I know that for $n=1$ it is trivial as the only group of order 1 is the trivial subgroup of $\{e\}$, but which groups should I consider for larger $n$ and how do I know I have considered all possibilities?

Michael Hardy
  • 1
Joald
  • 625

1 Answers1

3

For $n=4$, it is easy to check that the correct group is $$\mathbb{Z}_4\times\mathbb{Z}_2\ ,$$ which has order $8$.

There are two groups of order $6$: the cyclic group $\mathbb{Z}_6$ and the symmetric group $S_6$. The biggest cyclic subgroup of $S_6$ is a copy of $\mathbb{Z}_3$ (e.g. generated by $(123)$). It follows that the smallest group containing all the groups of order $6$ as subgroups cannot have order smaller than $12$. We take the group $$G=S_3\times\mathbb{Z}_2\ ,$$ then obviously $S_3$ is a subgroup, and we see $\mathbb{Z}_6$ as the subgroup generated by $((123),1)$. This group has order $12$.

The other cases are trivial.

Daniel Robert-Nicoud
  • 29,795
  • Any why is $n=6$ trivial? – freakish Aug 30 '17 at 11:50
  • @freakish Because there is only one group of order $6$. – Daniel Robert-Nicoud Aug 30 '17 at 11:51
  • What? There are two: $\mathbb{Z}_6$ and the symmetric group $S_3$. – freakish Aug 30 '17 at 11:52
  • @freakish Damn, I feel stupid now! Sorry, it's been a while since my group theory class. I'll edit. – Daniel Robert-Nicoud Aug 30 '17 at 11:53
  • For $n=6$ the answer is the Dihedral group $\mbox{Dih}_6\simeq S_3\times \mathbb{Z}_2$ which has 12 elements. But I only know this because I've read the classification of small groups (and their subgroups), see https://en.wikipedia.org/wiki/List_of_small_groups#List_of_small_non-abelian_groups – freakish Aug 30 '17 at 11:57
  • @freakish I got to the same conclusion, see the updated answer. – Daniel Robert-Nicoud Aug 30 '17 at 12:00
  • Your answer is correct and I get why it is correct by looking at the table of groups, but how does one check all the groups? Which groups do you look at? For example, when I considered $n=2$ I thought of $Z_2$ and $Z_3^*$, and it took me a bit of work to show that they are isomorphic. So how do I know what groups of a given order are? – Joald Aug 30 '17 at 12:04
  • 1
    @Joald For small groups, you can build all the possible Cayley tables for the given order: https://en.wikipedia.org/wiki/Cayley_table – Daniel Robert-Nicoud Aug 30 '17 at 12:13