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Say I want to consider every distinct abelian group of order

$10800=2^4\cdot3^3\cdot5^2$

up to isomorphism.

By the first Sylow theorem, I know that every group that has $p$ as a factor in its order, has a Sylow p-subgroup. I also know that a group of prime order is cyclic. (Correct so far?).

If I'm looking at a finitely generated group, say:

$A=\mathbb{Z}_{50}\times\mathbb{Z}_{6}\times\mathbb{Z}_{6}\times\mathbb{Z}_{6}$

Then this group has elementary divisors $2,2,2,2, 3,3, 3,5^2$.

Does this mean that A only has Sylow 2-subgroups of order 2? Or does it have Sylow 2-subgroups of order $2^4$?

And am I right in that only the Sylow 2-subgroups of order 2 are cyclic, since 2 to any other power is not prime?

Thanks in advance!

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If your group has order $2^43^35^2$ then the Sylow $2$-subgroups are order $2^4$. It's always the largest power of your prime that divides the order of the group.

You are also correct that a group of prime order is cyclic, so if $p$ was a prime and the largest power of $p$ that divided the order of your group was $p^1$ then the Sylow $p$-subgroups would indeed be cyclic.

Note this doesn't apply to any of the groups you've mentioned though, so their Sylow subgroups might be cyclic groups of order $p^n$ for whatever the appropriate $n$ is, or they might be some other non-cyclic group of order $p^n$.

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$2$-subgroups have orders that are a power of $2$, but Sylow $2$-subgroups have order the largest power of $2$ possible. That's $2^4$ in this case, like you pointed out. The Sylow theorems do indeed guarantee the existence of a subgroup of order $16$.

If $A = \Bbb Z_{50} \times \Bbb Z_6\times \Bbb Z_6\times \Bbb Z_6$, it's not too hard to find a group $P \leq A$ whose order is $16$. Take the subgroup isomorphic to $\Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_2$ that we can realize as $P = \langle (25, 0, 0, 0), (0, 3, 0, 0), (0, 0, 3, 0), (0, 0, 0, 3) \rangle$, for example.

And yes, groups of prime order are always cyclic, Sylow $p$-subgroup, or otherwise.

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