----
![[p-Sylow subgroup#^425d4e]]
> [!theorem] Theorem. ([[the Sylow theorems|first Sylow theorem]] )
> (Existence) $G$ contains a [[p-Sylow subgroup]].
> \
> ($p$-subgroups are nested) In turn, $G$ has a [[subgroup]] of [[order of a group|order]] $p^{k}$, for each $p^{k}$ [[divides|dividing]] $|G|$. Also, every [[p-group|p-subgroup]] with fewer than $p^{r}$ elements sits inside one of the larger [[p-group|p-subgroups]].
> ![[CleanShot 2023-09-22 at
[email protected]|300]]
^cd6a0a
> [!note]
> This is, in a sense, a generalization of [[Cauchy's Theorem]]:
> ![[CleanShot 2023-09-22 at
[email protected]]]
> [!theorem] Theorem. ([[the Sylow theorems|second Sylow theorem]], **weak version**)
> (Relationship) Any two [[p-Sylow subgroup|p-Sylow subgroups]] are [[conjugate]] (and hence [[group isomorphism|isomorphic]]) in $G$.
^a428b3
> [!theorem] ([[the Sylow theorems|third Sylow theorem]])
> (Counting) The number of [[p-Sylow subgroup|p-Sylow subgroups]] $G$ contains
> 1. [[divides]] $m$;
> 2. is [[congruent]] to $1 \text{ mod p}$.
**Remark.** Together, these imply $\text{\#(p-Sylow subgroups) | }m$.
Also, if we let $P$ denote a [[p-Sylow subgroup]] of $G$, then the number of [[p-Sylow subgroup|p-Sylow subgroups]] $G$ contains equals the [[index of a subgroup|index]] of the [[normalizer of a subgroup|normalizer]] of $P$ in $G$.
^457a0b
> [!basicexample] Example. (Classification of groups of order $15$)
(might be sus )
Of course, $G$ could equal $C_{15}$.
$15=3^{1} \cdot 5^{1}$. Letting $p=3$ and $m=5$, the [[the Sylow theorems|first sylow theorem]] implies $G$ has a $3$-[[p-Sylow subgroup|sylow subgroup]] of [[order of a group|order]] $3$. Letting $p=5$ and $m=3$, the [[the Sylow theorems|first Sylow theorem]] implies $G$ has a $5$-[[p-Sylow subgroup|sylow subgroup]] of order $5$. These [[subgroup]]s intersect trivially because [[greatest common divisor]]$(3,5)=1.$ So $C_{5} \times C_{3}$ is an option, if it isn't [[group isomorphism|isomorphic]] to $C_{15}$. There are $8$ elements left over in the group. By [[Lagrange's Theorem]], these elements have possible order $3$ or $5$ or $15$.
The only number that [[divides|divides]] $15$ and is [[congruent]] to $1 \text{ mod }3$ is $1$. The only number that [[divides]] $15$ and is [[congruent]] to $1 \text{ mod }5$ is $1$. Hence there are no other $3$ or $5$ sylow subgroups. So the other elements have order $15$, thus our group must be [[group isomorphism|isomorphic to]] $C_{15}$.
In fact, $G$ equals the [[internal direct product of subgroups|internal direct product]] of those [[subgroup]]s and therefore is [[linear isomorphism]] to $C_{5} \times C_{3}$.
> [!basicexample] Example. (Classifying groups of order 21)
> Let $G$ be a [[group]] with $|G|=21$.
\
$21=3^{1}\cdot 7^{1}$. The first and third Sylow theorems thus tell us that there exist either exactly $1$ or exactly $7$ $3$-sylow [[subgroup]]s of order $3$. They also tell us that there is exactly 1 $7$-sylow [[subgroup]].
\
We first ask ourselves: could $7$ 3-Sylow [[subgroup]]s of order $3$ even fit into $G?$ Note that they all must intersect trivially, since if their intersection ([[intersection of subgroups is a subgroup|which is a subgroup]]) were to have two elements then [[Lagrange's Theorem]] wouldn't hold. $1+6+2(7)=21$, so we can't *count them out*.
\
In the case that $G$ contains exactly one $3$-Sylow $K$ and exactly one $7$-Sylow $H$, we know that $G \cong C_{21}$. This is because $H$ and $K$ are both [[normal subgroup]]s (invariant under [[conjugate|conjugation]] by the second Sylow theorem), $H \cap K = (e)$, and $HK=G$ by [[cardinality of frobenius product of subgroups]], hence $H K$ is an [[internal direct product of subgroups|internal direct product]], [[group isomorphism|isomorphic]] to $C_{7} \times C_{3}$ which is in turn [[group isomorphism|isomorphic to]] $C_{21}$ since $3$ and $7$ are [[relatively prime integers|coprime]].
\
So assume the other case: that $G$ contains exactly $7$ $3$-Sylows. Fix one of them; call it $K:=\langle y \rangle$. Let $H=\langle x \rangle$ denote the $7$-Sylow as before. Since $H$ is a [[normal subgroup]] of $G$, $HK = KH$, and so by [[Frobenius product of subgroups|this characterization]] we know $HK \leq G$. ```tikz
\begin{document}
\begin{tikzpicture}[node distance=2cm]
% Nodes
\node (HK) {HK};
\node (H) [below left of=HK] {H};
\node (K) [below right of=HK] {K};
\node (e) [below of=H, node distance=2cm] {(e)};
% Lines connecting the nodes
\draw (HK) -- (H);
\draw (HK) -- (K);
\draw (H) -- (e);
\draw (K) -- (e);
\end{tikzpicture}
\end{document}```
In fact, the order of $HK$ must be $21$, so $HK=G$. Thus, if our prospective group exists it has the form $\{ hk:h \in H, k \in K \}=\{ x^{i}y^{j}: 0 \leq i \leq 6, 0 \leq j \leq 2 \}.$
Since $H$ is [[normal subgroup|normal]] in $G$, it is invariant under [[conjugate|conjugation]]. So take $x \in \langle x \rangle$ and [[conjugate]] it by $y$, and consider $yxy^{-1} \in H$. This element must have the form $x^{i}$ for some $i \in \{ 0,\dots,6 \}$. What choices of $i$ could work?
>- $i=0$: this doesn't work, because then $yxy^{-1}=e$, implying $x=e$.
>- $i=1$: this works, but it gives nothing new: $yxy^{-1}=x$ implies $yx=xy$, so $G$ is [[abelian group|abelian]], so all [[subgroup]]s are normal, and we are back in the $C_{3} \times C_{7} \cong C_{21}$ case.
>- $i\geq 2:$ Suppose $yxy^{-1}=x^{i}$. [[Conjugate]] both sides to get $y^{2}x^{}y^{-2}=yx^{i}y^{-1}=(yxy^{-1})^{i}=x^{{i}^{2}}$. Do it again to get $y^{3}x^{}y^{-3}=(yxy^{-1})^{i^{2}}=x^{i^{3}}$. But $y^{3}=e_{K}$, so we conclude that $x^{}=x^{i^{3}}$, i.e., that $1=i ^{3} \text{ mod } 7$. This is only true for $i \in \{ 1,2,4 \}$.
\
We know that if $i=1$ $G \cong C_{21}$; we now investigate $i=2$ and $i=4$. The claim is that they yield the same [[group isomorphism|isomorphism]] of $G$, and that [[group isomorphism|isomorphism]] is not between $G$ and $C_{21}$.
\
Let $i=2$, so that $yxy^{-1}=x^{2}$. Then multiplication is defined as $yx=x^{2}y$ and the group presentation is $\langle x,y \ | \ x^{7}=e, y^{3}=e, yx=x^{2}y \rangle .$
Let $i=4$, so that the group presentation is $\langle a,b \ | \ a^{7}=e, b^{3}=e, ba=a^{4}b \rangle .$
These groups are [[group isomorphism|isomorphic]] under the map $f:G_{i=2} \to G_{i=4}$ that sends $x \mapsto a^{2}$ and $y \mapsto b$. $f$ can be extended to a [[group homomorphism]] since $x$ and $y$ satisfy the same relations as $a$ and $b$. The inverse map is obtained by observing $a$ and $b$ satisfy the same relations as $x$ and $y$.
^64a314
> [!basicexample] Example. (Classification of order 30 groups)
Let $G$ be a [[group]] with $|G|=30$. To begin, note that $\begin{align}
30 \\
= & 2^{1} \cdot 15 \\
= & 3^{1} \cdot 10 \\
= & 5^{1} \cdot 6, \\
\end{align}$
and moreover $10 \equiv 1 \text{ mod }3$, $3 \equiv 1 \text{ mod 2}$, $5 \equiv 1 \text{ mod 2}$, and $15 \equiv 1 \text{ mod 2}$. From these facts we conclude by the [[the Sylow theorems|first and third Sylow theorems]] that $G$ could contain $\begin{align}
n_{2} & \in \{ 1,3,5,15 \} \ \ 2\text{-Sylow subgroups} \\
n_{3} & \in \{ 1,10 \} \ \ 3\text{-Sylow subgroups} \\
n_{5} & \in \{ 1,6 \} \ \ 5\text{-Sylow subgroup} \\
\end{align}$
We immediately see that $G$ has one [[normal subgroup]] $N_{5}$ of order $5$ or $K_{3}$ of order $3$. This is because by the [[the Sylow theorems|second Sylow theorem]], [[conjugate|conjugating]] a [[p-Sylow subgroup]] yields another [[p-Sylow subgroup]]— but there is exactly one [[p-Sylow subgroup]] of one of these orders (not enough space to have $n_{5}=6$ *and* $n_{3}=10$); said [[subgroup]] needs to be invariant under [[conjugate|conjugation]] by elements of $G$. Since $K_{3}$ and $N_{5}$ are both [[cyclic subgroup|cyclic]] of [[prime number|prime order]], $K_{3} \cap N_{5}=(e)$. Also, since $N_{5}$ or $K_{3}$ is a [[normal subgroup]] of $G$, $N_{5}K_{3} = K_{3}N_{5}$, and so by [[Frobenius product of subgroups|this characterization]] we know $N_{5}K_{3} \leq G$ [[cardinality of frobenius product of subgroups|with order]] $|N_{5}| |K_{3}|=15$. This product is a [[normal subgroup]] of $G$ since [[subgroups of index 2 are normal|it has index 2]]. Thus, $G$ must have a [[normal subgroup]] $N_{5}K_{3}$ of order $15$. Also, because $5$ and $3$ are [[relatively prime integers|coprime]], $N_{5}K_{3}$ is [[cyclic subgroup|cyclic]].
\
Notice that $N_{5}K_{3} \cap H_{2} = (e)$ for any $2$-Sylow $H$, for a nontrivial intersection would be a [[subgroup]] of order $2$ and $2 \not{|} 15$. $H_{2}$ is a [[normal subgroup]] of $G$ since it is [[cyclic group|abelian]]; hence $(N_{5}K_{3})H_{2}$ is a [[subgroup]] of $G$. In fact, by [[cardinality of frobenius product of subgroups]], $|(N_{5}K_{3})H_{2}|=30$, hence $G=(N_{5}K_{3})H_{2} \cong C_{15} \times C_{2}$.
\
Thus $G$ has the form $\{ x^{i}y^{j}: 0 \leq i \leq 14, 0 \leq j \leq 1 \}.$
What can the multiplication $yx$ look like on $G$? $N_{5}K_{3}$ is [[normal subgroup|normal]], hence $yxy^{-1}=x^{i}$ for some $i \in \{ 0,\dots1 4\}$. What choices of $i$ are valid?
We note that $\begin{align}
x=y^{2}xy^{-2}=yx^{i}y^{-1}=(yxy^{-1})^{i}=x^{i^{2}},
\end{align}$
so it must be true that $1=i^{2} \text{ mod } 15$. This is true for all $i \in \{ 1, 4,11, 14 \}$. So the possible presentations for $G$ are $\begin{align}
G & = \langle x,y \ | \ x^{15}=e, y^{2}=e, yx=xy \rangle \\
G & = \langle x,y \ | \ x^{15}=e, y^{2}=e, yx=x^{4}y \rangle \\
G & = \langle x,y \ | \ x^{15}=e, y^{2}=e, yx=x^{11}y \rangle \\
G & = \langle x,y \ | \ x^{15}=e, y^{2}=e, yx=x^{14}y \rangle.
\end{align}$
>This means that, at most, there are $4$ groups of order $30$. Here are $4$ groups of order 30 which we know are mutually non-[[linear isomorphism]]:
>- $C_{30}$
>- $D_{15}$
>- $D_{5} \times C_{3}$
>- $D_{3} \times C_{5}$
^4691c0
> [!basicexample]
> Let $M$ be a 'mystery group' with $|M|=200$. How much of $M