MHB Problem of the Week #123 - August 4th, 2014

[Chris L T521]

Here's this week's problem!

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Problem: Let $p$ and $q$ be distinct primes. How many mutually non-isomorphic Abelian groups are there of order $p^2q^4$?

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[Chris L T521]

This week's problem was correctly answered by Deveno. You can find his solution below.

[sp]Let $|G| = p^2q^4$. If we set $P$ as the Sylow $p$-subgroup of $G$, and $Q$ as the Sylow $q$-subgroup of $G$, then since $G$ is Abelian, both of these subgroups are normal with $G = PQ$ and $P \cap Q = \{e\}$.

Thus $G \cong P \times Q$.

It therefore suffices to determine the isomorphism types of $P$ and $Q$, which are themselves Abelian groups.

Now $P \cong \Bbb Z_{p^2}$ or $P \cong \Bbb Z_p \times \Bbb Z_p$ (depending on whether or not $P$ has an element of order $p^2$).

Similarly (using, for example, the Structure Theorem for Finitely-Generated Abelian Groups):

$Q \cong \Bbb Z_{q^4}$
$Q \cong \Bbb Z_{q^3} \times \Bbb Z_q$
$Q \cong \Bbb Z_{q^2} \times \Bbb Z_{q^2}$
$Q \cong \Bbb Z_{q^2} \times \Bbb Z_q \times \Bbb Z_q$ or
$Q \cong \Bbb Z_q \times \Bbb Z_q \times \Bbb Z_q \times \Bbb Z_q$ (these correspond to the distinct partitions of 4)

Thus we have 2 possible choices for $P$ and 5 possible choices of $Q$, giving 10 possible non mutually non-isomorphic Abelian groups in all.[/sp]