Sylow subgroups exist

Statement

Let $G$ be a finite group and $p$ be a prime number. Then, there exists a $p$-Sylow subgroup $P$ of $G$: a subgroup whose order is a power of $p$ and whose index is relatively prime to $p$.

Note that when $p$ does not divide the order of $G$, the $p$-Sylow subgroup is trivial, so the statement gives interesting information only when $p$ divides the order of $G$.

This statement is often viewed as a part of a more general statement called Sylow's theorem.

Related facts

Other parts of Sylow's theorem

• Sylow implies order-dominating: Given any $p$-Sylow subgroup and any $p$-subgroup, the $p$-Sylow subgroup contains a conjugate of the $p$-subgroup.
• Sylow implies order-conjugate: All $p$-Sylow subgroups are conjugate. This follows directly from the previous part.
• Congruence condition on Sylow numbers: The number of $p$-Sylow subgroups is congruent to $1$ modulo $p$.

Stronger forms of existence

• Every Sylow subgroup intersects the center nontrivially or is contained in a centralizer: Every Sylow subgroup of a group either has a nontrivial intersection with the center of the group, or it is contained in the centralizer of some non-central element.

Analogues for Hall subgroups

The analogous statement does not hold for Hall subgroups. A Hall subgroup is a subgroup whose order and index are relatively prime. A $\pi$-Hall subgroup is a Hall subgroup such that the set of primes dividing its order is contained in $\pi$, and the set of primes dividing its index is disjoint from $\pi$.

• Hall subgroups need not exist: Given a set $\pi$ of primes, there may not exist a $\pi$-Hall subgroup.
• Hall subgroups exist in finite solvable group: If, however, the finite group is solvable, then it has $\pi$-Hall subgroups for all prime sets $\pi$.
• Hall's theorem: This states that a finite group is solvable if and only if it has $\pi$-Hall subgroups for every prime set $\pi$.

Analogues in other algebraic structures

• Sylow subloops exist for Sylow primes in finite Moufang loop
• 2-Sylow subloops exist in finite Moufang loop
• 3-Sylow subloops exist in finite Moufang loop
• Hall subloops exist in finite solvable Moufang loop

Facts used

1. Lucas' theorem (more specifically, Lucas' theorem prime power case)
2. Fundamental theorem of group actions: There is a bijection between the coset space of the stabilizer of an element and the orbit of that element. In particular, the size of the orbit equals the index of the stabilizer.
3. Lagrange's theorem
4. Class equation of a group
5. Cauchy's theorem for abelian groups
6. Central implies normal: Any subgroup inside the center is normal.

Proof

This proof uses a tabular format for presentation. Provide feedback on tabular proof formats in a survey (opens in new window/tab) | Learn more about tabular proof formats|View all pages on facts with proofs in tabular format

Proof by action on subsets

Given: A finite group $G$ of order $n=p^{r}m$ where $p$ is prime, $r$ is a nonnegative integer, and $p$ does not divide $m$.

To prove: $G$ has a subgroup of order $p^r$.

Proof: Here are some observations regarding this action:

Step no.	Construction/assertion	Facts used	Previous steps used	Explanation
1	Let $W$ be the set of all subsets of $G$ of size $p^r$.
2	The size of $W$ is relatively prime to $p$	Fact (1)		The size of $W$ is the binomial coefficient $(\genfrac{}{}{0ex}{}{n}{p^r})$, and an appeal to Lucas's theorem (fact (1)) reveals that its value is relatively prime to $p$.
3	$G$ acts on $W$ by left multiplication.			First, note that a group acts on the set of all its subsets by left multiplication. Further, since the left multiplication maps are bijective, they preserve the sizes of subsets. In particular, any subset of size $p^r$ gets mapped to a subset of size $p^r$. Thus, we can restrict the action to the set $W$ of subsets of size $p^r$.
4	The action of $G$ on $W$ has at least one orbit, say $\mathcal{O}$, whose size is relatively prime to $p$.		Step (2)	The size of $W$ equals the sum of sizes of the orbits under the $G$-action. Since the total is relatively prime to $p$, at least one of the numbers has to be relatively prime to $p$.
5	Let $S$ be a member of such an orbit $\mathcal{O}$ and $H$ be the stabilizer of $S$ in $G$.
6	The index of $H$ in $G$ equals the size of $\mathcal{O}$.	Fact (2)	Steps (4), (5)	Follows directly from fact (2).
7	The index of $H$ in $G$ is relatively prime to $p$.		Steps (4), (6)	By step (4), the size of $\mathcal{O}$ is relatively prime to $p$. By step (6), this equals the index of $H$. Thus, the index of $H$ in $G$ is relatively prime to $p$.
8	The index of $H$ in $G$ divides $m$.	Fact (3)	Step (6)	By fact (3) (Lagrange's theorem), the index of $H$ in $G$ divides the order of $G$, which is $p^{r}m$. By step (6), the index of $H$ in $G$ is relatively prime to $p$. Hence, it must divide $m$.
9	The index of $H$ in $G$ is at least $m$.		Steps (1), (4), (5)	The union of $gS,g\in G$, is the whole group $G$. Since each $gS$ has size $p^r$, there are at least $n/p^r=m$ translates of $S$ in $G$. Thus, $\mathcal{O}$ has size at least $m$.
10	The index of $H$ in $G$ is exactly $m$, so $H$ is a $p$-Sylow subgroup of $G$.		Steps (8), (9)	The index of $H$ in $G$ divides $m$ and is at least $m$. The only way both of these could be true is if it equals $m$.

Interestingly, the only nontrivial result we use here (Lucas' theorem) can itself be proved using group theory by doing the above proof in reverse taking a cyclic group (although a purely algebraic proof also exists). ,

Proof by conjugation action

Given: A finite group $G$ of order $n=p^{r}m$, where $p$ is prime, $r$ is a nonnegative integer, and $p$ does not divide $m$.

To prove: $G$ has a subgroup of order $p^r$.

Proof: We prove the claim by induction on the order of $G$. Specifically, we assume that the result is true for all groups of order smaller than the order of $G$.

For the case $r=0$, the proof is direct since the trivial group is a subgroup of order $p^r$. Thus, we assume $r>0$. In particular, $p$ divides the order of $G$.

Consider the class equation of $G$ (fact (4)):

$\left|G\right|=|Z(G)|+\sum _{i=1}^r|G:C_G(g_i)|$

where $c_1,c_2,\dots ,c_r$ are the conjugacy classes of non-central elements and $g_i$ is an element of $c_i$ for each $i$.

We consider two cases:

Case that $p$ divides the order of $Z(G)$ :

Step no.	Construction/assertion	Facts used	Previous steps used	Explanation
1	There exists a normal subgroup of order $p$ in $G$	Facts (5), (6)		Since $Z(G)$ is Abelian, fact (5) yields that it has a subgroup $H$ of order $p$. Since $H$ is in the center, $H$ is normal in $G$ (by fact (6)). Thus, $H$ is a normal subgroup of $G$ of order $p$.
2	$G/H$ has a $p$-Sylow subgroup, in particular, a subgroup $Q$ of order $p^{r-1}$		Step (1)	Since $H$ has order $p$, $G/H$ has order $p^{r}m/p=p^{r-1}m$. This is strictly smaller than the order of $G$, so induction applies and we get a $p$-Sylow subgroup, whose order is $p^{r-1}$.
3	Let $\alpha :G\to G/H$ be the quotient map. Then ${\alpha }^{-1}(Q)$ is a $p$-Sylow subgroup of $G$		Steps (1), (2)	Indeed, the order of ${\alpha }^{-1}(Q)$ is the product of the order of $H$ and the order of $Q$, which is therefore $p\cdot p^{r-1}=p^r$.

Case that $p$ does not divide the order of $Z(G)$

Step no.	Construction/assertion	Facts used	Previous steps used	Explanation
1	There exists $i$ such that $p$ does not divide the index $k$ of $C_G(g_i)$ in $G$			Since $p$ divides the order of $G$, $p$ cannot divide the index of every $C_G(g_i)$, otherwise the class equation would yield that $p$ divides the order of $Z(G)$.
2	$C_G(g_i)$ is a proper subgroup of $G$ whose order is a multiple of $p^r$, so $p^r$ is the largest power of $p$ dividing its order.	Fact (3)	Step (1)	Since $g_i$ is non-central, $C_G(g_i)$ is proper in $G$. Further, since $|G:C_G(g_i)|=k$ is relatively prime to $p$, Lagrange's theorem (fact (3)) yields that the order of $C_G(g_i)$ is $p^{r}m/k$, which is a multiple of $p^r$.
3	$C_G(g_i)$ contains a subgroup of order $p^r$		Step (2)	This follows by the induction hypothesis.
4	$G$ contains a subgroup of order $p^r$, and hence a $p$-Sylow subgroup.		Step (3)	A subgroup of order $p^r$ in $C_G(g_i)$ is also a subgroup of order $p^r$ in $G$.