[Math] The number of cyclic subgroups of order 15 in $\mathbb{Z}_{30} \oplus \mathbb{Z}_{20}$

abstract-algebracyclic-groupsfinite-groupsgroup-theory

What is the number of cyclic subgroup of order 15 in $\mathbb{Z}_{30} \oplus \mathbb{Z}_{20}$?

I have counted the number of elements with order 15 in $\mathbb{Z}_{30} \oplus \mathbb{Z}_{20}$ that is $48$, by counting all possibilities of $(a,b)$ where $a\in\mathbb{Z}_{30}$ , $b \in \mathbb{Z}_{20}$ and $\def\lcm{\operatorname{lcm}}\lcm(|a|,|b|)=15$.

My argument is that for an element that has order $15$, say $(x,y)$, then

$\langle(x,y)\rangle = \langle(x,y)^2\rangle = \langle (x,y)^4\rangle = \langle (x,y)^7\rangle = …=\langle (x,y)^{14}\rangle$

So there are $8$ different elements with order $15$ such that the generate the same subrgroup. So my answer is that there are $\frac{48}{8}=6$ different cyclic subgroups of order 15,
am i correct? or maybe there are more simpler ways to solve this problem?

Best Answer

Provided you correctly counted the elements of order$~15$, your answer is correct. You can indeed count cyclic subgroups by counting their generators (elements or order$~n$) and dividing by the number $\phi(n)$ of generators per cyclic subgroup, since every element of order$~n$ lies in exactly one cyclic subgroup of order$~n$ (the one that it generates).

Here is how I would count the elements of order$~15$. By the Chinese remainder theorem one has $\newcommand\Z[1]{\Bbb Z_{#1}}\Z{30}\cong\Z2\oplus\Z3\oplus\Z5$ and $\Z{20}\cong\Z4\oplus\Z5$, so all in all we are dealing with the group $\Z2\oplus\Z4\oplus\Z3\oplus\Z5^2$. To have order $15$, an element must have a trivial (zero) component in $\Z2$ and $\Z4$, in $\Z3$ it must have as component one of the $2$ generators, and it's component in $\Z5^2$ must be any one of the $24$ nonzero elements. Indeed you get $2\times24=48$ elements of order$~15$.

Related Solutions

[Math] How Many Cyclic Subgroups of order $10$ are there in $\mathbb{Z}_{100}\oplus\mathbb{Z}_{25}$

By using GAP 4.6.5 these desired subgroup are enclosed computationally. Below is needed codes for doing this job:

gap> f:=FreeGroup("a","b");;
     a:=f.1;; b:=f.2;;
     s:=f/[a^100,b^25,a*b*a^(-1)*b^(-1)];;
     e:=AllSubgroups(s);;
     c:=Filtered(e,t->IsCyclic(t)=true);;
     Filtered(c,t->Order(t)=10);

 Group([ b^5, a^-50 ]),
 Group([ a^-9*b^10*a^-1 ]),
 Group([ a^-9*b^-10*a^-1 ]),
 Group([ a^8*b^-5*a^2 ]),
 Group([ a^8*b^5*a^2 ]), 
 Group([ a^-10 ]) ]

[Math] Finding Number of Cyclic Sub groups of order $15$ in $Z_{30} \oplus Z_{20}$. The mistake in this method

First off, the easy way to do such an exercise is to first use the Chinese remainder theorem ($\def\Z#1{\mathbf Z_{#1}}\Z{kl}\cong \Z k\oplus\Z l$ when $\gcd(k,l)=1$) to put your group into better shape. Here one has $$\Z{30}\oplus\Z{20}\cong(\Z2\oplus\Z3\oplus\Z5)\oplus(\Z4\oplus\Z5)\cong(\Z2\oplus\Z4)\oplus\Z3\oplus\Z5^2,$$ and any cyclic group of order $15$ must project to the three factors on the right giving respectively the trivial group (there are no elements of even order), the full group $\Z3$, and a cyclic subgroup (of $\Z5^2$) of order $5$. Moreover the cyclic subgroup will be the direct sum of its projections, so it is entirely determined by those projections. Only the final projection leaves any choice, and there are $6$ choices possible (the six one-dimensional subspaces of a vector space of dimension $2$ over the field $\Z5$).

Now honestly, I don't understand all of what you are doing in the question. Are you counting subgroups of order $15$ or elements $(g,h)$ of order $15$? Maybe you are counting subgroups by trying to understand where such elements can be found, but you should be more clear about that. Anyway you seem to be missing an important case, namely where $|g|=15$ and $|h|=5$. That case accounts for four of the six subgroups, while the two cases you enumerate each only account for one subgroup. In your cases the subgroup is determined by the subgroups $\langle g\rangle$ and $\langle h\rangle$ of the two summands (because it is the sum of those subgroups) and $\langle g\rangle$ and $\langle h\rangle$ are unique since any cyclic subgroup contains at most one subgroup of any given order. This explains why you found only two subgroups.

However in the missing case, just knowing $\langle g\rangle$ and $\langle h\rangle$ does not suffice, as various choices are possible for $g,h$ that give these subgroups. One must also know exactly which $g$ pairs up with which $h$, and this can be done so as to obtain $4$ different subgroups of order $15$.

So this may raise a question: if a subgroup of a sum of two cyclic groups does not have to be the sum of its projections onto those groups, why was it possible to reason with only the three projections for the sum $(\Z2\oplus\Z4)\oplus\Z3\oplus\Z5^2$ in the first paragraph? This is so because there the summands are Abelian groups of pairwise relatively prime order. In such a situation any subgroup must contain its projections on the summands, and therefore be the sum of those projections; this exercise should make you appreciate the power of that statement. Here is why it is true: the Chinese remainder theorem implies that for each summand there is some integer $k$ such that multiplication by $k$ amounts on that summand (by reduction modulo the order of the summand) to multiplication by $1$, while for all other summands it amounts to multiplication by $0$. Then multiplication by $k$ realises the projection onto that summand, but on the other hand it is an operation that maps any subgroup into itself.

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Related Solutions

[Math] How Many Cyclic Subgroups of order $10$ are there in $\mathbb{Z}_{100}\oplus\mathbb{Z}_{25}$

[Math] Finding Number of Cyclic Sub groups of order $15$ in $Z_{30} \oplus Z_{20}$. The mistake in this method

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