## Tag Archives: sylow subgroup

### Prove that a given tensor product of ZZ/(pⁿ) by a finite abelian group A is isomorphic to the Sylow p-subgroup of A

Let $A$ be a finite abelian group of order $n$ and let $p^k$ be the largest power of the prime $p$ dividing $n$. Prove that $\mathbb{Z}/(p^k) \otimes_\mathbb{Z} A$ is isomorphic to the Sylow $p$-subgroup of $A$.

We saw in this previous exercise that as a $\mathbb{Z}$-module, $A$ is the internal direct sum of its Sylow subgroups; say $A = \bigoplus_Q M_q$, where $Q$ is the (finite) set of primes dividing $n$. Now tensor products commute with finite direct sums, so that $\mathbb{Z}/(p^k) \otimes_\mathbb{Z} A \cong_\mathbb{Z} \mathbb{Z}/(p^k) \otimes_\mathbb{Z} (\bigoplus_Q M_q) \cong_\mathbb{Z} \bigoplus_Q (\mathbb{Z}/(p^k) \otimes_\mathbb{Z} M_q)$.

We claim that if $q \neq p$, then $\mathbb{Z}/(p^k) \otimes_\mathbb{Z} M_q = 0$. To see this, let $|M_q| = q^\ell$. Since $q$ and $p$ are relatively prime in the principal ideal domain $\mathbb{Z}$, there exist integers $a$ and $b$ such that $aq^\ell + bp^k = 1$. Now let $\overline{x} \otimes y$ be a simple tensor in $\mathbb{Z}/(p^k) \otimes_\mathbb{Z} M_q$. We have $\overline{x} \otimes y = (\overline{x} \cdot 1) \otimes y$ $= \overline{x} \cdot (aq^\ell + bp^k) \otimes y$ $= \overline{x} \cdot aq^\ell \otimes y + \overline{x} \cdot bp^k \otimes y$ $= \overline{x} \cdot a \otimes q^\ell \cdot y + \overline{ap^k} \cdot b \otimes y$ $= 0 + 0 = 0$. Certainly then every element in this tensor product is zero, so that $\mathbb{Z}/(p^k) \otimes_\mathbb{Z} M_q = 0$.

Next, we claim that $\mathbb{Z}/(p^k) \otimes_\mathbb{Z} M_p \cong_\mathbb{Z} M_p$. To see this, note first that every simple tensor (hence every element) in $\mathbb{Z}/(p^k) \otimes_\mathbb{Z} M_p$ can be written in the form $\overline{1} \otimes m$, since for an arbitrary simple tensor we have $\overline{a} \otimes b = (\overline{1} \cdot a) \otimes b$ $= \overline{1} \otimes a \cdot b$. Now define $\varphi : \mathbb{Z}/(p^k) \times M_p \rightarrow M_p$ by $\varphi(\overline{a},m) = a \cdot m$. Note that if $p^k$ divides $a-b$, then (a-b) \cdot m = 0\$, so that $a \cdot m = b \cdot m$. In particular, $\varphi$ is well-defined. Evidently, $\varphi$ is $\mathbb{Z}$-balanced, and so induces a group homomorphism $\Phi : \mathbb{Z}/(p^k) \otimes_\mathbb{Z} M_p \rightarrow M_p$. $\Phi$ is surjective since $\otimes{1} \otimes m \mapsto m$. Since $M_p$ and $\mathbb{Z}/(p^k) \otimes_\mathbb{Z} M_p$ are finite sets, $\Phi$ is a bijection and thus a group isomorphism.

Thus we have $\mathbb{Z}/(p^k) \otimes_\mathbb{Z} A \cong_\mathbb{Z} M_p$.

### A finite group whose order is the product of three distinct primes has a normal Sylow subgroup of largest order

Let $G$ be a group of order $pqr$ where $p < q < r$ and $p$, $q$, and $r$ are primes. Prove that a Sylow $r$-subgroup of $G$ is normal.

Recall that some Sylow subgroup of $G$ is normal. Let $P$, $Q$, and $R$ denote Sylow $p$-, $q$-, and $r$-subgroups of $G$, respectively.

Suppose $n_p(G) = 1$. Note that $n_r(G/P) \in \{1,q\}$, and since $q < r$, $n_r(G/P) = 1$. Let (via the lattice isomorphism theorem) $\overline{R} \leq G$ be the subgroup whose quotient is the unique Sylow $r$-subgroup in $G/P$; we have $|\overline{R}| = pr$, and $P \leq \overline{R}$ is normal. Moreover, because $p < r$, $\overline{R}$ has a unique Sylow $r$-subgroup $R$, and we have $\overline{R} \cong P \times R$. Now if $R^\prime \leq G$ is a Sylow $r$-subgroup, then $R^\prime P/P = \overline{R}/P$, so that $\overline{R} = R^\prime P$. Since $R^\prime \leq \overline{R}$ is Sylow, $R^\prime = R$. Thus $n_r(G) = 1$.

The same argument works if $n_q(G) = 1$.

### Normal subsets of a Sylow subgroup which are conjugate in the supergroup are conjugate in the Sylow normalizer

Let $A$ and $B$ be normal subsets of a Sylow $p$-subgroup $P \leq G$. Prove that if $A$ and $B$ are conjugate in $G$ then they are conjugate in $N_G(P)$.

[With generous hints from Wikipedia.]

Suppose $B = gAg^{-1}$, with $g \in G$. Now since $P \leq N_G(A)$, $gPg^{-1} \leq N_G(gAg^{-1}) = N_G(B)$.

Note that $P, gPg^{-1} \leq N_G(B)$ are Sylow, and thus are conjugate in $N_G(B)$. So for some $h \in N_G(B)$, we have $P = hgPg^{-1}h^{-1}$, so that $hg \in N_G(P)$. Moreover, $hgAg^{-1}h^{-1} = hBh^{-1} = B$. Thus $A$ and $B$ are conjugate in $N_G(P)$.

### The normalizer of a maximal Sylow intersection is not a p-subgroup

Suppose that over all pairs of Sylow $p$-subgroups, $P$ and $Q$ are chosen so that $|P \cap Q|$ is maximal. Prove that $N_G(P \cap Q)$ is not a $p$-group.

We proved this in the course of answering this previous exercise.

### A sufficient condition for Sylow intersections of bounded index

Prove that if $n_p \not\equiv 1$ mod $p^k$, then there are distinct Sylow $p$-subgroups $P$ and $Q$ in $G$ such that $P \cap Q$ has index at most $p^{k-1}$ in both $P$ and $Q$.

(We will follow the strategy of this previous exercise.)

Let $P \leq G$ be a Sylow $p$-subgroup. Now $P$ acts on $\mathsf{Syl}_p(G)$ by conjugation; note that if $Q \in \mathsf{Syl}_p(G)$ is distinct from $P$, then using the Orbit-Stabilizer theorem and Lemma 4.19, we have $|P \cdot Q| = [P : N_P(Q)]$ $= [P : P \cap N_G(Q)]$ $= [P : Q \cap P]$. Since the orbit containing $P$ itself has order 1, then $n_p(G) = 1 + \sum [P : Q_i \cap P]$, where $Q_i$ ranges over a set of orbit representatives. If every $[P : Q_i \cap P]$ is divisible by $p^k$, then $n_p(G) \equiv 1$ mod $p^k$, a contradiction. Thus some $[P : Q_i \cap P]$ is a power of $p$ and is at most $p^{k-1}$.

We can see then that $Q_i \cap P$ has index at most $p^{k-1}$ in both $P$ and $Q_i$.

### Every group of order 36 has a normal Sylow subgroup

Prove that if $G$ is a group of order 36, then $G$ has either a normal Sylow 2-subgroup or a normal Sylow 3-subgroup.

Note that $36 = 2^2 \cdot 3^2$, so that Sylow’s Theorem forces $n_2(G) \in \{1,3,9\}$ and $n_3(G) \in \{1,4\}$.

Suppose $n_3(G) \neq 1$. Then $n_3(G) = 4 \not\equiv 1$ mod 9, so that by Lemma 13 and this previous exercise, there exist $P_3,Q_3 \in \mathsf{Syl}_3(G)$ such that $P_3 \cap Q_3$ is nontrivial. Consider now $C_G(P_3 \cap Q_3)$; since $P_3$ and $Q_3$ are abelian, $\langle P_3,Q_3 \rangle$ centralizes $P_3 \cap Q_3$, and moreover we can see by Sylow’s Theorem that $G = \langle P_3, Q_3 \rangle$. Thus $P_3 \cap Q_3 \leq Z(G)$, and in fact $P_3 \cap Q_3$ is the intersection of all Sylow 3-subgroups of $G$. Thus $G$ contains $2 + 6 + 6 + 6 + 6 = 26$ elements of 3-power order. Let $P_2 \leq G$ be a Sylow 2-subgroup; $P_2$ contains three nonidentity elements, whose products with the nonidentity elements in $P_3 \cap Q_3$ yield 6 elements of order 6 or 12. There are 4 elements left, which necessarily comprise a unique Sylow 2-subgroup.

### The normalizer of a maximal Sylow p-intersection contains more than one Sylow p-subgroup

Let $G$ be a group with more than one Sylow $p$-subgroup. Over all pairs of distinct Sylow $p$-subgroups let $P$ and $Q$ be chosen so that $|P \cap Q|$ is maximal. Show that $N_G(P \cap Q)$ has more than one Sylow $p$-subgroup and that any two distinct Sylow $p$-subgroups of $N_G(P \cap Q)$ intersect in $P \cap Q$. (Thus $|N_G(P \cap Q)|$ is divisible by $p|P \cap Q|$ and by some prime other than $p$. Note that Sylow $p$-subgroups of $N_G(P \cap Q)$ need not by Sylow in $G$.)

Note that $P \cap Q < N_P(P \cap Q)$ is proper since $P$ is a $p$-group. Since $N_P(P \cap Q) \leq N_G(P \cap Q)$, we have that $p|P \cap Q|$ divides $|N_G(P \cap Q)|$.

Suppose now that $N_G(P \cap Q)$ is a $p$-group. Then $N_G(P \cap Q) \leq R$ for some Sylow $p$-group $R$. If $R = P$, then $N_G(P \cap Q) \leq N_P(P \cap Q)$, and hence $N_G(P \cap Q) = N_P(P \cap Q)$. However, since $P \cap Q < N_Q(P \cap Q) \leq N_G(P \cap Q) \leq P$ and $P \cap Q < N_Q(P \cap Q) \leq Q$, we have a contradiction since there are elements in $P$ and $Q$ but not in $P \cap Q$ (namely, $N_Q(P \cap Q) \setminus P \cap Q$). Thus $R \neq P$. But then we have $P \cap Q < N_P(P \cap Q)$ $\leq N_G(P \cap Q)$ $\leq R$ and $P \cap Q < N_P(P \cap Q) \leq P$, so that $P \cap Q < N_P(P \cap Q) \leq P \cap R$. This implies that $P \cap Q$ is not (cardinality) maximal among the pairwise intersections of Sylow $p$-subgroups. Thus $N_G(P \cap Q)$ is not a $p$-group, so that its order is divisible by some prime other than $p$.

Suppose now that $N_G(P \cap Q)$ contains a unique Sylow $p$-subgroup, say $R$. Then $N_P(P \cap Q) \leq R$. Moreover, $R \leq P^\prime$ for some Sylow $p$-sugbroup $P^\prime$ of $G$. Since $P \cap Q < N_P(P \cap Q) \leq P \cap P^\prime$ and $P \cap Q$ is order maximal, we in fact have $R \leq P^\prime = P$. Similarly, $R \leq Q$. But then $P \cap Q \leq R \leq P \cap Q$, so that $P \cap Q$ is Sylow in $N_G(P \cap Q)$, a contradiction.

Let $A, B \in \mathsf{Syl}_p(N_G(P \cap Q))$. Since $P \cap Q$ is a $p$-group, $P \cap Q$ is contained in some Sylow $p$-group $R \leq N_G(P \cap Q)$. By Sylow’s Theorem, $R$ is conjugate to $A$ and $B$ in $N_G(P \cap Q)$, say by $x$ and $y$, respectively. Since $P \cap Q$ is normal in $N_G(P \cap Q)$, $P \cap Q \leq A, B$. Thus $P \cap Q \leq A \cap B$. Now $A \leq A^\prime$ and $B \leq B^\prime$ for some Sylow $p$-subgroups $A^\prime$ and $B^\prime$ of $G$. Suppose $A^\prime = B^\prime$. Then, since $A$ and $B$ are distinct, the subgroup $\langle A. B \rangle$ as generated in $A^\prime$ is a $p$-group, normalizes $P\cap Q$, and properly contains (for instance) $A$, which is a contradiction since $A$ is Sylow in $N_G(P \cap Q)$. Thus we have $A^\prime \neq B^\prime$. Moreover, $A \cap B \leq A^\prime \cap B^\prime$, and since $|P \cap Q|$ is maximal among the intersections of distinct Sylow $p$-subgroups and $G$ is finite, $P \cap Q = A \cap B$.

### If any Sylow p-subgroup intersects all others trivially, then all Sylow p-subgroups intersect trivially

Let $G$ be a finite group and $p$ a prime. Suppose that for some $P \in \mathsf{Syl}_p(G)$, we have $P \cap R = 1$ for all $R \in \mathsf{Syl}_p(G)$ with $R \neq P$. Prove that $P_1 \cap P_2 = 1$ for all $P_1,P_2 \in \mathsf{Syl}_p(G)$. Deduce that in this case, the number of nonidentity elements of $p$-power order in $G$ is precisely $(|P|-1)[G:N_G(P)]$.

Let $P_1$ and $P_2$ be distinct Sylow $p$-subgroups. Now by Sylow’s Theorem, $P_1$ is conjugate to $P$ in $G$; say $xP_1x^{-1} = P$. Now if we also have $xP_2x^{-1} = P$, then $P_1 = x^{-1}Px = P_2$, a contradiction. Note then that $x(P_1 \cap P_2)x^{-1} = xP_1x^{-1} \cap xP_2 x^{-1}$ $= P \cap Q$, where $Q$ is some Sylow $p$-subgroup of $G$ distinct from $P$. Thus $x(P_1 \cap P_2)x^{-1} = 1$, and we have $P_1 \cap P_2 = 1$.

Now every nonidentity element of $p$-power order is in some Sylow $p$-subgroup of $G$, an no such element is in more than one. The number of nonidentity elements in a Sylow $p$-subgroup is $|P|-1$ and the number of Sylow $p$-subgroups is $n_p(G) = [G:N_G(P)]$, and the final conclusion follows.

### The Frattini subgroup of a finite group is nilpotent

Let $G$ be a finite group. Prove that $\Phi(G)$ is nilpotent.

We will use Frattini’s Argument to show that every Sylow subgroup is normal.

Let $P \leq \Phi(G)$ be a Sylow subgroup. By Frattini’s Argument, $G = \Phi(G) N_G(P)$. By a lemma to a previous theorem, $N_G(P)$ cannot be proper, so that $N_G(P) = G$ and we have $P \leq G$ normal. Then $P \leq \Phi(G)$ is normal; thus all of the Sylow subgroups of $\Phi(G)$ are normal, and thus $\Phi(G)$ is nilpotent.

### Classification of groups of order 28

Classify the finite groups of order 28.

We begin with a lemma.

Lemma 1: Let $H$ and $K$ be groups, $\varphi, \psi : K \rightarrow \mathsf{Aut}(H)$ group homomorphisms, and $\theta : K \rightarrow K$ an automorphism. If $\psi = \varphi \circ \theta$, then the mapping $\Phi : H \rtimes_\varphi K \rightarrow H \rtimes_\psi K$ given by $\Phi((h,k)) = (h,\theta^{-1}(k))$ is an isomorphism. Proof: (Homomorphism) Let $(h_1,k_1), (h_2,k_2) \in H \rtimes_\varphi K$. Then $\Phi((h_1,k_1)(h_2,k_2)) = \Phi((h_1 \varphi(k_1)(h_2), k_1k_2))$ $= (h_1 \varphi(k_1)(h_2), \theta^{-1}(k_1k_2))$ $= (h_1 \varphi(\theta(\theta^{-1}(k_1)))(h_2), \theta^{-1}(k_1) \theta^{-1}(k_2))$ $= (h_1 (\varphi \circ \theta)(\theta^{-1}(k_1))(h_2), \theta^{-1}(k_1) \theta^{-1}(k_2))$ $= (h_1 \psi(\theta^{-1}(k_1))(h_2), \theta^{-1}(k_1) \theta^{-1}(k_2))$ $= (h_1, \theta^{-1}(k_1))(h_2, \theta^{-1}(k_2))$ $\Phi((h_1,k_1)) \Phi((h_2,k_2))$. (Injective) If $\Phi((h_1,k_1)) = \Phi((h_2,k_2))$, then $(h_1,\theta^{-1}(k_1)) = (h_2,\theta^{-1}(k_2))$, so that $h_1 = h_2$ and $k_1 = k_2$. (Surjective) If $(h,k) \in H \rtimes_\psi K$, then $(h,k) = \Phi((h, \theta(k)))$. $\square$

Now to the main result.

Note that $28 = 2^2 \cdot 7$.

By FTFGAG, there are two distinct abelian groups of order 28: $Z_{28}$ and $Z_{14} \times Z_2$.

Suppose now that $G$ is a nonabelian group of order 28. By Sylow’s Theorem, we have $n_7 = 1$; thus $G$ has a unique (hence normal) Sylow 7-subgroup $H \cong Z_7$. Let $K \leq G$ be a Sylow 2-subgroup. By Lagrange, we have $H \cap K = 1$; thus $G = HK$, and by the recognition theorem for semidirect products, $G \cong H \rtimes_\varphi K$ for some $\varphi : K \rightarrow \mathsf{Aut}(H)$. Evidently, classifying the nonabelian groups of order 28 is equivalent to classifying the distinct groups constructed in this way. To that end, let $H = Z_7 = \langle y \rangle$. Then $\mathsf{Aut}(Z_7) \cong Z_6 = \langle \alpha \rangle$, where $\alpha(x) = x^2$.

Suppose $K = Z_4 = \langle x \rangle$. Now every homomorphism $\varphi : Z_4 \rightarrow Z_6$ is determined by $\varphi(x)$, and moreover, provided $\varphi(x)^4 = 1$, any map thus defined is indeed a homomorphism. Thus we have two choices for $\varphi(x)$: $1$ and $\alpha^3$. If $\varphi(x) = 1$, then $\varphi$ is trivial, contradicting the nonabelianicity of $G$. Thus $\varphi(x) = \alpha^3$ is the only group homomorphism $K \rightarrow \mathsf{Aut}(H)$ which gives rise to a nonabelian group $Z_7 \rtimes_\varphi Z_4$ of order 28.

Suppose now that $K = Z_2^2 = \langle a \rangle \times \langle b \rangle$. Every homomorphism $\varphi : K \rightarrow \mathsf{Aut}(H)$ is determined uniquely by $\varphi(a)$ and $\varphi(b)$, and provided $\varphi(a)^2 = \varphi(b)^2 = 1$, every mapping thus defined is a homomorphism. Thus we have $\varphi(a), \varphi(b) \in \{1, \alpha^3 \}$, for a total of four choices.

If $\varphi_1(a) = \varphi_1(b) = 1$, then $\varphi$ is trivial, contradicting the nonabelianicity of $G$.

If $\varphi_2(a) = \alpha^3$ and $\varphi_2(b) = 1$, then $Z_7 \rtimes_{\varphi_2} Z_2^2$ is indeed a nonabelian group of order 28.

If $\varphi_3(a) = 1$ and $\varphi_3(b) = \alpha^3$, then we have $\varphi_3 = \varphi_2 \circ \theta$, where $\theta(a) = b$ and $\theta(b) = a$. By the Lemma, $H \rtimes_{\varphi_3} K \cong H \rtimes_{\varphi_2} K$. Likewise, if $\varphi_4(a) = \alpha^3$ and $\varphi_4(b) = \alpha^3$, we do not get an essentially new group by the lemma, via the map $\theta(a) = a$ and $\theta(b) = ab$.

Thus there is a unique nonabelian group of order 28 whose Sylow 2-subgroups are not cyclic.

Thus, the distinct groups of order 28 are as follows. In all cases, $Z_7 = y$, $Z_4 = \langle x \rangle$, and $Z_2^2 = \langle a \rangle \times \langle b \rangle$.

1. $Z_{28}$
2. $Z_{14} \times Z_2$
3. $Z_7 \rtimes_\varphi Z_4$ where $\varphi(x)(y) = y^{-1}$
4. $Z_7 \rtimes_\psi Z_2^2$ where $\psi(a)(y) = y^{-1}$ and $\psi(b)(y) = y$