Question

How many Sylow 5-subgroups, Sylow 3-subgroups and Sylow 2-subgroups can a
group of order 200 have? Give reasons for your answers.

Answer 1

Answer on Question #81169 – Math – Abstract Algebra

Question

How many Sylow 5-subgroups, Sylow 3-subgroups and Sylow 2-subgroups can a group of order 200 have? Give reasons for your answers.

Solution

Case p=3: Firstly, we write $|G| = 200 = 2^3 5^2$. As order 3 does not divide order of G, there is no element of order 3. There is no 3-Sylow subgroup.

Case p= 5: Let P be a Sylow 5-subgroup of G. It exists by Sylow Theorem 1.

Let $N_G(P)$ be a normalizer of P and $n_S$ be the number of 5-Sylow subgroups in G.

By Sylow Orbit-Stabilizer Theorem, $n_S = |G: N_G(P)|$.

By Sylow Theorem 3, $(n_S - 1) \vdots 5$.

As $P \leq N_G(P), |N_G(P)| \vdots |P|$.

So, 25 divides $N_G(P)$ and possible values of $|G: N_G(P)| = |G|: |N_G(P)| = 200: |N_G(P)|$ are 1, 2, 4, 8.

The only possibility is $n_S = 1$. By the way, that provide as with $N_G(P) = G$ and P is normal in G.

Theorem (Sylow III). For each prime p, let n_p be the number of p-Sylow subgroups of G, $|G| = p^k m = 5^2 2^3$, where p=5 doesn't divide m=8. Then n_p=1 mod p=1 mod 5 and n_p | m, n_p | 8.

The only possibility for n_p=1 mod 5 and n_p | 8 is n_p=1. There is one 5-Sylow subgroup.

Case p=2:

Let Q be a Sylow 2-subgroup of G and $n_2$ number of 2-Sylow subgroups in G.

By Sylow Orbit-Stabilizer theorem, $n_2 = |G: N_G(Q)|$.

As $Q \leq N_G(Q), |N_G(Q)| \vdots 8$. Thus, possible values for $n_2$ are 1, 5, 25.

Construction for $n_2 = 1$: If G any Abelian group, i.e. cyclic group, then $n_2 = 1$.

Construction for $n_2 = 25$: Consider Dihedral group $D_{25}$, the group of symmetries of a regular 25-gon.

In a way similar as above, we can show that its Sylow 5-subgroup is normal. Hence, it is unique (corollary Sylow Theorem 2). Let's consider remaining elements. Their orders cannot have 5 as a divisor as all such elements are in a unique 5-Sylow subgroup. Thus, their order is 2. There are

$|D_{2S}| - 25 = 25$ such elements. So, they are all conjugate to each other (also Sylow Theorem 2). So, there are 25 Sylow 2-subgroups in $D_{2S}$.

Finally, consider $G = D_{2S} \times \mathbb{Z}_4$. If $P_2$ is a 2-Sylow subgroup in $D_{2S}$, $R := P_2 \times \mathbb{Z}_4$ is 2-Sylow subgroup in $G$. Cardinal of orbit of $R$ in $G$ is the same as the cardinal of orbit of $P_2$ in $D_{2S}$. This is true as the direct product of $D_{2S}$ and $\mathbb{Z}_4$. Thus, $n_2 = 25$ for $G$.

Construction for $n_2 = 5$: $G = D_5 \times \mathbb{Z}_{20}$. Similar to case above, we show that $D_5$ has 5 Sylow 2-subgroups and so has $G$.

**Answer**: there is no 3-Sylow subgroup, there is only one 5-Sylow subgroup and the number of 2-Sylow subgroups is one of numbers in the set $\{1, 5, 25\}$.

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