The number of abelian groups up to isomophism of order $6^5$ is
$a)\ 5\\ b)\ 7\\ c)\ 49\\ d)\ 65$
Asked
Active
Viewed 98 times
2 Answers
0
I think the answer is $49$...$6^5=2^5 3^5$ and number of distinct partition of $5 =7$...so there are $7(7) =49$ number of abelian group up to isomorphism of order $6^5$.
Juniven Acapulco
- 9,654
Halima.Khatun
- 807
-
You are correct. But do you understand why you can use partitions? – Alex Macedo Dec 30 '16 at 06:29
0
well, it is $P(5)^2$.
We calculate $P(n)$ with the euclidean recurrence starting with $P_0$:
$1,1,2,3,5,7$.
So the answer is $7^2$
Asinomás
- 105,651
-
We don't really have to calculate anything. 49 is the only square number among the answer options, so... – Ivan Neretin Dec 30 '16 at 06:32
-
-
-
-
by the fundamental theorem of abelian groups the group is a product of prime power cyclic subgroups, so we can treat the powers of $2$ and $3$ seperately, for the powers of $2$ you need groups of size $2^a$ such that the sum of the exponents is $5$, there are $P(5)$ ways to do so, the same happens for the powers of $3$. – Asinomás Dec 30 '16 at 06:47