We need to find the Number of Cyclic Sub groups of order $15$ in $Z_{30} \oplus Z_{20}$ . This method does not give me the right answer (i.e $6$ ) .
Attempt: We need to find the number of Cyclic Sub groups of order $15$ in $Z_{30} \oplus Z_{20}$ . Consider $(g,h) \in Z_{30} \oplus Z_{20}$.
Since $(g,h)$ is of order $15 => lcm(|g|, |h|) = 15$
CASE 1 : $ |g| = 3 , |h| =5 $
CASE 2 : $|g| = 15 , |h| =1$
CASE $1$
Since, $Z_{30}$ and $Z_{20}$ are cyclic groups. And cyclic groups have unique subgroups.
Hence, When $|g|= 3$, $g$ can be be any element of order $3$ from $\langle 10 \rangle \in Z_{30}$ ;
$ g \in \langle 10 \rangle = \{10,20,0\} ~s.t. |g| =3$. We see that $10, 20$ both have order $= 3$ . Check .: $\phi (3) = 2$
$\langle 10 \rangle$ is the unique cyclic sub group of order $3$ in $Z_{30}$. Hence, if an element of order $3$ exists in $Z_{30}$, it must exist in $\langle 10 \rangle$ only.
We also know that If $G$ and $H$ are cyclic subgroups then, $G \oplus H$ is cyclic $<=> |G|,|H|$ are prime to each other.
Hence, we can select only that sub group of $Z_{20}$ whose order is prime to $3$. Thus, we will select that sub group of $Z_{20}$ whose order is $5$.
Such a sub group in $Z_{20}$ is unique and is $\langle 4 \rangle = \{4,8,12,16,0\}$ . We see that the elements $\{4,8,12,16\}$ all are of order $5$ .. Check : . $\phi(5) = 4$
Hence, PICK any $g \in \{10,20\}$ and any $h \in \{4,8,12,16\}$ : $g \oplus h$ yields one cyclic sub group of order $15$.
CASE $2$ : $|g| = 15 , |h| =1$
In this case, we can select $g$ from $\langle 2 \rangle \leq Z_{30} =\{2,4,6,8,……,26,28,0\}$ . We have $ \phi(15) = 8$ such elements who have order as $15$ for $g$ and one element in $Z_{20}$ which has order $1$.
hence, $g \oplus h$ yields one more cyclic sub group of order $15$. In total, there should be only $2$ sub group of order $15$ of $Z_{30} \oplus Z_{20}$ .
EDIT : But the answer is $6$ by an another method. There are $48$ elements of order $15$ in $Z_{30} \oplus Z_{20}$ .If $a \in Z_{30}$, then $a$ can assume orders $\{1,2,3,5,6,10,15,30\}$ and if $b \in Z_{20}$, then $b$ can assume orders $1,2,4,5,10,20$ by Lagrange's Theorem.
Then, since, $\phi (3) = 2, \phi (5) = 4, \phi(15) = 8$ , number of elements of order $15 = 2.4 + 8.1 + 8.4 = 48$. Since, in any cyclic group, the number of elements of order $15=\phi(15) = 8$ . hence, number of cyclic sub groups = $48/8=6$.
. Where am i making a mistake? Help will be really appreciated. Thank you
Best Answer
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.