Difference between definitions of $p$-subgroup and Sylow $p$-subgroup

Question

I'm reading Abstract Algebra by Dummit and Foote and the following definitions are made:

$1$. A group of order $p^{\alpha}$ for some $\alpha\geq1$ is called a $p$-group. Subgroups of $G$ which are $p$-groups are called $p$-subgroups.

$2$. If $G$ is a group of order $p^{\alpha}m$, where $p$ does not divide $m$, then a subgroup of order $p^{\alpha}$ is called a Sylow $p$-subgroup of G.

As I see it the difference seems to be that if $H \leq G$ with $|H|=p^{\beta}$ for $\beta\geq1$ it is called a $p$-subgroup if $|G|=p^{\alpha}$ and a sylow subgroup of $G$ if its order is not a power of $p$.

But in the wording of the Sylow's theorem in the book it reads that

Let G be a group of order $p^{\alpha}m$, where $p$ is a prime not dividing $m$ then...If $P$ is a Sylow $p$-subgroup of $G$ and $Q$ is any $p$-subgroup of G...

I don't understand the difference between $P$ and $Q$..both are subgroups of $G$ with order of a power of $p$ and I thought that $Q$ can't be called a $p$ subgroup because the order of $G$ is not a power of $p$.

So it seems that I didn't understand the difference in definitions $(1)$ and $(2)$ after all.

Can someone please clarify ?

Answer 1

A $p$-group is a group where the order of every element is a power of $p$. A direct consequence of this is that if the order of every element is a power of $p$, then the order of the group must be a power of $p$. This follows from Lagrange's theorem: the order of a subgroup divides the order of the group. So, one can say that a $p$-group is a group with order a power of $p$.

A Sylow $p$-subgroup, also called a $p$-Sylow subgroup, is a maximal $p$-subgroup of a group. This means that it is not a proper subgroup of any other $p$-subgroup of the main group. In other words, if $G$ is a group with order $q^{\alpha}m$, where $q \nmid m$, then the first Sylow theorem guarantees us a subgroup with order $q^{\alpha}$. This subgroup is called a Sylow $q$-subgroup of $G$.

One can say that every Sylow $p$-subgroup is a $p$-group, but not every $p$-group is a Sylow $p$-subgroup.

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Let's take a group $G$ with order $45$. Then $\left|G\right|=45=3^2 \cdot 5$.

A subgroup of $G$ with order $3^2$ is a Sylow $3$-subgroup. A subgroup of $G$ with order $5$ is a Sylow $5$-subgroup of $G$. However, a subgroup with order $3$ is a $3$-subgroup, but it is NOT a Sylow $3$-subgroup.