Elements and cyclic subgroups of order $15$ in $\Bbb Z_{30}\times \Bbb Z_{20}.$

alternative-proofcombinatoricscyclic-groupsfinite-groupsgroup-theory

This is Exercise 8.22 of Gallian's, "Contemporary Abstract Algebra". Please use only methods from this book prior to the exercise.

This is an question.

Find the number of elements of order $15$ and the number of cyclic subgroups of order $15$ in $\Bbb Z_{30}\times \Bbb Z_{20}.$

Thoughts:

This exercise gives two distinct tasks, yes, but I think it's to highlight the discrepancy between the two answers.

I've done some work below in GAP for this but I don't consider it a solution within the spirit of the text.

gap>  F:=FreeGroup(2);
<free group on the generators [ f1, f2 ]>
gap> 
gap> rels:=[(F.1)^(30), (F.2)^(20), (F.1)*(F.2)*(F.1)^-1*(F.2)^-1];
[ f1^30, f2^20, f1*f2*f1^-1*f2^-1 ]
gap> G:=F/rels;
<fp group on the generators [ f1, f2 ]>

This sets the group up, calling it G.

gap> S:=[];
[  ]
gap> 
gap> for g in G do
> if Order(g)=15 then
> AddSet(S, g);
> fi;
> od;
gap> 
gap> 
gap> Size(S);
48

So I think there are $48$ elements of order $15$ in G.

gap> C:=[];
[  ]
gap> 
gap> for g in G do
> H:=Subgroup(G, [g]);
> if Size(H)=15 then
> AddSet(C, H);
> fi;
> od;
gap> 
gap> 
gap> Size(C);
6

So I think there are $6$ cyclic subgroups of G of order $15$.

That's all I have.

I think I should be able to do this exercise myself but I lack the time to take it any further without help.

Here is a similar question I asked: Is $|\{z\in\Bbb Z_3\times\Bbb Z_9: |z|=9\}|=18?$

Please help 🙂

Best Answer

I don't have this text, so I don't know which material is available to you, but the following only uses basic facts about cyclic groups and coprimality.

Hint Since $\Bbb Z_{20} \cong \Bbb Z_4 \times \Bbb Z_5$ and $\Bbb Z_{30} \cong \Bbb Z_2 \times \Bbb Z_3 \times \Bbb Z_5$, we have $$\Bbb Z_{20} \times \Bbb Z_{30} \cong \Bbb Z_2 \times \Bbb Z_3 \times \Bbb Z_4 \times \Bbb Z_5^2 .$$

Since the group is abelian, any element of order $15$ is a product of an element of order $3$ and an element of order $5$; conversely, any such product has order $15$ and any two such products are distinct. How many elements of order $3$ are there in the above product? Of order $5$?

With that number in hand, each element of order $15$ is contained inside exactly one subgroup of order $15$ (namely, the one it generates). How many elements of order $15$ are in each such subgroup?

From the above decomposition we see that there are two elements of order $3$ and $24$ elements of order $5$, so there are $2 \cdot 24 = 48$ elements of order $15$. Now, any subgroup of order $15$ contains $\phi(15) = 8$ elements of order $15$, and any element of order $15$ is in precisely one subgroup of that order (namely the one it generates), so the $48$ elements of order $15$ are evenly divided across $\frac{48}{8} = 6$ subgroups of that order.

Related Question