4
$\begingroup$

Can you provide some examples of Sylow $p$-subgroups in infinite groups?

(Sylow $p$-subgroups are $p$-subgroups that are maximal with respect to inclusion among all $p$-subgroups in the group.)

$\endgroup$
  • 6
    $\begingroup$ Presumably you want a single conjugacy class of such subgroups, since the existence of maximal $p$-subgroup does not seem to be asking much. Upper unitriangular matrices in ${\rm GL}(n,k)$ is one standard example, where $k$ is algebraicaly closed of prime characteristic $p.$ Amalgamated products of finit groups provide another set of examples, if certain compatibility conditions are met. $\endgroup$ – Geoff Robinson Dec 15 '11 at 22:12
  • 4
    $\begingroup$ The Prufer $p$-group is its own Sylow $p$-subgroup, and it's the Sylow $p$-subgroup of $\mathbb{Q}/\mathbb{Z}$. $\endgroup$ – Arturo Magidin Dec 16 '11 at 1:57
  • $\begingroup$ Incidentally, if we take a free product of two cyclic groups of order $p,$ for any prime $p,$ then we get a group which has precisely two conjugacy classes of maximal $p$-subgroups. $\endgroup$ – Geoff Robinson Dec 17 '11 at 12:13
5
$\begingroup$

Here's a very simple example of a group which has two maximal $p$-subgroups which are not conjugate. In fact, it's a special case of the comment I made. Let $G = \langle a,b : a^2 = abab = 1 \rangle.$ Then $G$ is an infinite dihedral group with an infinite cyclic normal subgroup $B = \langle b \rangle.$ it is easy to check that every element of the coset $aB$ has order $2,$ whereas all non-identity elements of $B$ have infinite order. Hence all maximal $2$-subgroups of $G$ have order $2.$ We claim that $a$ and $ab$ are not conjugate in $G.$ For, given any integer $j,$ we have $(ab^j)^{-1}a(ab^j) = b^{-j}ab^j = aab^{-j}ab^j = ab^{j}b^{j} = ab^{2j}.$ Hence we see that all elements of the form $ab^{2j}$ are conjugate to $a,$ and similarly (or just allowing $ab$ to play the role of $a),$ we see that all elements of the form $ab^{2j+1}$ are conjugate to $ab.$

$\endgroup$
2
$\begingroup$

I'll give you some examples.

You can consider $\bigoplus_{n=0}^\infty \mathbb Z/p\mathbb Z$, it is a $p$-Sylow by itself, or if you prefer something which isn't a $p$-group by itselfs can consider the group $\mathbb Z \oplus \bigoplus_{n=0}^\infty \mathbb Z/p\mathbb Z$ other.

Another example could be $\mathbb Z \oplus \mathbb Z/p\mathbb Z$, in this case the $p$-Sylow is finite.

As you can see you can create as many example as you wish.

$\endgroup$
  • $\begingroup$ The tensor product distributes over direct sums, so $$\mathbb{Z}\otimes\bigoplus_{n=0}^{\infty}\mathbb{Z}/p\mathbb{Z}\cong \bigoplus_{n=0}^{\infty}(\mathbb{Z}\otimes\mathbb{Z}/p\mathbb{Z}) \cong \bigoplus_{n=0}^{\infty}\mathbb{Z}/p\mathbb{Z},$$so your "second" example is exactly the same as the first and is a $p$-group. $\endgroup$ – Arturo Magidin Dec 16 '11 at 20:28
  • $\begingroup$ Sorry, I've mistaken the simbol, obviously I meant $\oplus$ not $\otimes$, I've corrected, thanks for the correction. $\endgroup$ – Giorgio Mossa Dec 17 '11 at 22:57
1
$\begingroup$

Maximal $p$-subgroups can become quite ugly in infinite groups, even if the groups are locally finite (i.e., each finitely generated subgroup is finite).

Take for example a look at the construction in:

Marianna Dalle Molle, Sylow Subgroups Which Are Maximal in the Universal Locally Finite Group of Philip Hall, Journal of Algebra 215, 229-34 (1999)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy