Tag Archives: subgroup lattice

Prove that two given groups have isomorphic subgroup lattices

Let p be an odd prime, let P_1 = Z_p \times Z_p^2, and let P_2 be the nonabelian group of order p^3 which has an element of order p^2. Prove that P_1 and P_2 have isomorphic subgroup lattices.


The nontrivial subgroups of Z_p \times Z_{p^2} = \langle x \rangle \times \langle y \rangle have order p or p^2. By §6.1 #28, we know that P_1 has precisely p+1 subgroups of order p^2. By this previous exercise, P_1 also has precisely p+1 subgroups of order p.

Note that \langle x \rangle \times \langle y^p \rangle \cong Z_p \times Z_p is maximal in P_1 and elementary abelian. We claim that this is the unique elementary abelian maximal subgroup of P_1. To see this, note that P_1 contains (p+1)(p-1) = p^2-1 elements of order p. Including the identity, there are exactly p^2 elements x in P_1 such that x^p=1. The set of these elements is a subgroup, and moreover any other elementary abelian subgroup of rank 2 must coincide with this one. In particular, one maximal subgroup is isomorphic to Z_p \times Z_p and the rest must be cyclic, by the classification of groups of order p^2. Label the elementary abelian maximal subgroup Q and the cyclic maximal subgroups R_1, \ldots, R_p. Each of the R_i contains a unique subgroup of order p, while Q contains p+1 such subgroups- that is, Q contains all of the order p subgroups of P_1. Moreover, we see that the order p subgroup of each R_i is \Phi(P); that is, R_i \cap R_i = \Phi(P) for all distinct i and {j}. Label the non-Frattini order p subgroups T_i for 1 \leq i \leq p.

The subgroup lattice of P_1 is thus as follows.

  1. There is a unique subgroup of order p^3; namely P_1.
  2. There are p+1 subgroups of order p^2; namely Q and R_i, 1 \leq i \leq p. Each of these is contained in P_1.
  3. There are p+1 subgroups of order p; namely \Phi(P) and T_i, 1 \leq i \leq p. Moreover, \Phi(P) is contained in every maximal subgroup, while each T_i is contained only in Q.
  4. There is a unique subgroup of order 1; namely 1. This subgroup is contained in each subgroup of order p.

Now we consider the group P_2. By this previous exercise, we know that P_2 has precisely p+1 subgroups of order p^2. By this previous exercise, P_2 has a subgroup isomorphic to Z_p^2; label this subgroup Q.

We claim that Q is the unique elementary abelian maximal subgroup of P_2. To see this, we will find a presentation for P_2 and prove some basic facts using it.

From the discussion on page 183 in the text on order p^3 groups, we see that P_2 has the presentation \langle x, y \ |\ x^{p^2} = y^p = 1, yx = x^{1+p}y \rangle. Clearly every element of this group can be written in the form x^iy^j, with 0 \leq i < p^2 and 0 \leq j < p. There are p^3 such expressions, and since we know that P_2 has order p^3, every element can be written in this form uniquely. We now prove a lemma.

Lemma: Let G be a group. If x,y \in G such that x^{p^2} = y^p = 1 and yx = x^{1+p}y. Then for all nonnegative integers a,b,n, we have (x^ay^b)^n = x^{na + n(n-1)abp/2} y^{nb}. Proof: If b = 0, the result is clear. Suppose b = 1; we proceed by induction on n. for the base case, note that (x^ay)^0 = 1 = x^{0a + 0}y^{0 \cdot 1}. Similarly, if n=1 we have (x^ay)^1 = x^{1 \cdot a + 0}y, and if n=2, then (x^ay)^2 = x^ayx^ay = x^ax^{a(1+p)}y^2 = x^{2a + 2 \cdot 1 \cdot ap/2}y^2. For the inductive step, suppose the conclusion holds for some n \geq 2. Then (x^ay)^{n+1} = (x^ay)^n(x^ay) = x^{na + n(n-1)ap/2}y^{n}x^ay = x^{na + n(n-1)ap/2}x^{a(1+p)^n}y^{n+1}. Since n \geq 2, and x^{p^2} = 1, all but the first two terms in the expansion of (1+p)^n are zero, so that (1+p)^n \equiv 1+np mod p^2. (Note that we used a lemma from this previous exercise.) Thus this is equal to x^{(n+1)a + (n+1)nap/2}y^{n+1}, and the conclusion holds for b=1. Suppose now that b \geq 2; we proceed by induction on n. If n=0, then (x^ay^b)^0 = 1 = x^{0a + 0}y^{0b}. For the inductive step, suppose the conclusion holds for some n \geq 0. Then (x^ay^b)^{n+1} = (x^ay^b)^n(x^ay^b) = x^{na + n(n-1)abp/2}y^{nb}x^ay^b = x^{na + n(n-1)abp/2}x^{a(1+p)^{nb}}y^{(n+1)b}. Again, all but the first two terms in the expansion of (1+p)^{nb} are 0 mod p^2, so that this element in fact equals x^{na + n(n-1)abp/2}x^{a(1+nbp)}y^{(n+1)b} = x^{(n+1)a + (n+1)nabp/2}y^{(n+1)b}. Thus the conclusion holds for all n. \square

We can now show that P_2 contains exactly p^2 elements g such that g^p = 1. If x^ay^b is arbitrary in P_2, then by the lemma (x^ay^b)^p = x^{ap}. (Here we use the fact that p is odd.) Thus we must have p|a. Thus if g^p = 1, a \equiv 0 mod p and b is unrestricted; there are p^2 such elements, and clearly every such element has order p (or 1). Thus P_2 has exactly p^2 elements of order p or 1, which necessarily comprise Q. Moreover, we see that any other elementary abelian subgroup of order p^2 is in fact equal to Q. Thus the remaining maximal subgroups are cyclic; label these R_1 through R_p.

Now \Phi(P_2) is an order p subgroup, and is the intersection of all pairs of maximal subgroups. \Phi(P) is in fact the unique order p subgroup in each of the R_i; any other order p subgroups are thus contained in Q (any two disjoint order p subgroups have an elementary abelian product). Q has p+1 such subgroups; thus P_2 has p+1 order p subgroups.

Thus we see that P_2 and P_1 have isomorphic subgroup lattices.

Exhibit the Sylow 3-subgroups of Alt(4) and Sym(4)

Exhibit all Sylow 3-subgroups of A_4 and S_4.


In this previous exercise we found the subgroup lattice of A_4; from this we can easily see that the Sylow 3-subgroups of A_4 are \langle (1\ 2\ 3) \rangle, \langle (1\ 2\ 4) \rangle, \langle (1\ 3\ 4) \rangle, and \langle (2\ 3\ 4) \rangle.

Now the Sylow 3-subgroups of S_4 have order 3; thus every nonidentity element in a Sylow 3-subgroup is a 3-cycle. Thus every Sylow 3-subgroup of S_4 is contained in A_4, and the Sylow 3-subgroups of S_4 are precisely those of A_4.

Distinct 3-cycles in Sym(4) which are not mutual inverses generate Alt(4)

Prove that if x and y are distinct 3-cycles in S_4 with x \neq y^{-1}, then \langle x, y \rangle = A_4.


In a previous exercise, we computed the subgroup lattice of A_4. From this lattice it is clear that the join of any two distinct cyclic subgroups of order 3 is all of A_4.

Alt(4) is generated by any pair of elements of order 2 and 3 respectively

Prove that a subgroup of A_4 generated by any element of order 2 and any element of order 3 is all of A_4.


In a previous exercise, we computed the subgroup lattice of A_4. From this lattice it is clear that the join of any cyclic subgroup of order 2 with any cyclic subgroup of order 3 is all of A_4.

Compute the subgroup lattice of Alt(4)

Find the subgroup lattice of A_4.


Note that A_4 \leq S_4 consists of 12 elements: the identity, eight 3-cycles, and the three products of two 2-cycles. All nonidentity elements of A_4 have prime order, so the minimal subgroups are \langle (1\ 2)(3\ 4) \rangle, \langle (1\ 3)(2\ 4) \rangle, \langle (1\ 4)(2\ 3) \rangle, \langle (1\ 2\ 3) \rangle, \langle (1\ 2\ 4) \rangle, \langle (1\ 3\ 4) \rangle, and \langle (2\ 3\ 4) \rangle.

Note that (a\ b)(c\ d) \circ (a\ c)(b\ d) = (a\ c)(b\ d) \circ (a\ b)(c\ d) = (a\ d)(b\ c) has order 2. Thus the join of any two subgroups generated by products of 2-cycles contains the third and has order 4.

Suppose H \leq A_4 contains a product of 2-cycles and a 3-cycle; without loss of generality, (a\ b)(c\ d), (a\ b\ c) \in H for some a,b,c,d. Now 1, (a\ b\ c), (a\ c\ b), (a\ b)(c\ d), (a\ b\ c)(a\ b)(c\ d) = (a\ c\ d), (a\ d\ c), and (a\ b)(c\ d)(a\ b\ c) = (b\ d\ c) are in H, so H has at least 7 elements. By Lagrange’s Theorem, H = S_4.

Now consider the join H of two subgroups generated by 3-cycles- without loss of generality, \langle (a\ b\ c) \rangle and \langle (a\ b\ d) \rangle. Now (a\ b\ c)(a\ b\ d) = (a\ c)(b\ d) \in H, so that by the previous paragraph, H = A_4.

Thus the subgroup lattice of A_4 is as follows. (Click for full size.)

Compute the subgroup lattice of a quotient of the modular group of order 16

Let M = \langle u,v \rangle be the modular group of order 16 described in a previous exercise. Prove that \langle v^4 \rangle is normal in M and use the lattice isomorphism theorem to draw the lattice of subgroups of M/\langle v^4 \rangle. Which group of order 8 has the same subgroup lattice? Use generators and relations to determine the isomorphism type of M/\langle v^4 \rangle.


Note that vv^4v^{-1} = v^4 \in \langle v^4 \rangle and uv^4u^{-1} = uv^4u = u^2v^{20} = v^4 \in \langle v^4 \rangle; by a previous exercise, \langle v^4 \rangle is normal in M.

By the Lattice isomorphism theorem, the subgroup lattice of M/\langle v^4 \rangle is as follows.

The group \mathbb{Z}/(2) \times \mathbb{Z}/(4) has the same subgroup lattice, and in fact M/\langle v^4 \rangle = \langle \overline{u}, \overline{v} \ |\ \overline{u}^2 = \overline{v}^4 = 1, \overline{v}\overline{u} = \overline{u} \overline{v} \rangle, so that M/\langle v^4 \rangle \cong \mathbb{Z}/(2) \times \mathbb{Z}/(4).

Draw the subgroup lattice of a quotient of the quasidihedral group of order 16

Let QD_{16} = \langle \sigma, \tau \rangle be the quasidihedral group of order 16 described in a previous exercise. Prove that \langle \sigma^4 \rangle is normal in QD_{16} and use the Lattice Isomorphism Theorem to draw the lattice of subgroups of QD_{16}/\langle \sigma^4 \rangle. Which group of order 8 has the same subgroup lattice? Use generators and relations to determine the isomorphism type of QD_{16}/\langle \sigma^4 \rangle.


Note that \sigma \sigma^4 \sigma^{-1} = \sigma^4 \in \langle \sigma^4 \rangle and \tau \sigma^4 \tau^{-1} = \tau \sigma^3 \sigma \tau = \sigma \tau \tau \sigma^3 = \sigma^4 \in \langle \sigma^4 \rangle. By a previous exercise, then, \langle \sigma^4 \rangle is normal in QD_{16}.

The subgroup lattice of QD_{16}/\langle \sigma^4 \rangle is as follows.

The dihedral group D_8 has the same subgroup lattice. Moreover, we have QD_{16}/\langle \sigma^4 \rangle = \langle \overline{\sigma}, \overline{\tau} \ |\ \overline{\sigma}^4 = \overline{\tau}^2 = 1, \overline{\sigma} \overline{\tau} = \overline{\tau} \overline{\sigma}^{-1} \rangle, so that QD_{16}/\langle \sigma^4 \rangle \cong D_8.

Compute the subgroup lattice of Q(8)

Compute the lattice of subgroups of Q_8.


Every \subseteq-minimal subgroup of Q_8 is cyclic; disregarding inverses and the identity, there are at most 4 distinct cyclic subgroups of Q_8: \langle -1 \rangle, \langle i \rangle, \langle j \rangle, and \langle k \rangle. Note that i^2 = j^2 = k^2 = -1, so that \langle -1 \rangle is a subgroup of the other three; by Lagrange’s Theorem, there are no intermediate subgroups. Thus \langle -1 \rangle is the only \subseteq-minimal subgroup of Q_8.

Now by Lagrange’s Theorem, any subgroup of Q_8 must have order 1, 2, 4, or 8. A subgroup of order 1 is trivial and a subgroup of order 2 is cyclic. If H \leq Q_8 is minimally generated by 2 elements, then those elements are (without loss of generality) (i,j), (i,k), or (j,k). In any case we have H = Q_8. We remark that any subgroup minimally generated by n > 2 elements contains Q_8 as a subgroup and hence is all of Q_8. Thus the only order 4 subgroups are cyclic, and we enumerated these above. Lastly, Q_8 itself is the only order 8 subgroup.

Thus the subgroup lattice of Q_8 is as follows.

The subgroup lattice of a symmetric group on three elements

Find the lattice of subgroups of S_3.


Every (nontrivial) \subseteq-minimal subgroup of S_3 is cyclic – i.e., generated by a single nonidentity element. There are 5 such subgroups: \langle (1\ 2) \rangle, \langle (1\ 3) \rangle, \langle (2\ 3) \rangle, \langle (1\ 2\ 3) \rangle, and \langle (1\ 3\ 2) \rangle. Note that \langle (1\ 2\ 3) \rangle = \langle (1\ 3\ 2) \rangle since (1\ 2\ 3)^{-1} = (1\ 3\ 2). Moreover, \langle (1\ 2\ 3) \rangle has order 3 and is thus distinct from the other cyclic subgroups, which have order 2. Finally, the order 2 cyclic subgroups are distinct because their generators are distinct. No order 2 subgroup is contained in \langle (1\ 2\ 3) \rangle by Lagrange’s Theorem, so that the distinct nontrivial \subseteq-minimal subgroups of S_3 are \langle (1\ 2) \rangle, \langle (1\ 3) \rangle, \langle (2\ 3) \rangle, and \langle (1\ 2\ 3) \rangle. Clearly any pairwise meet of these subgroups is trivial.

Now every other subgroup of S_3 is a finite join of some nontrivial \subseteq-minimal subgroups. Let H and K be two such subgroups. If |H| = 2 and |K| = 3, then \mathsf{lcm}(2,3) = 6 must divide |\langle H, K \rangle| by Lagrange’s Theorem; thus, since S_3 is finite, \langle H,K \rangle = S_3. If |H| = |K| = 2, then \mathsf{lcm}(2,2) = 2 must divide |\langle H,K \rangle| by Lagrange’s Theorem. If |\langle H,K \rangle| = 2, then \langle H,K \rangle has no proper subgroups and thus is \subseteq-minimal; that is, \langle H,K \rangle is one of the order 2 subgroups accounted for above, and thus H = K. If |\langle H,K \rangle| = 4, we have a contradiction by Lagrange’s Theorem. If |\langle H,K \rangle| = 6, then since S_3 is finite we have \langle H,K \rangle = S_3. We have seen that the pairwise join of two distinct \subseteq-minimal subgroups of S_3 is all of S_3. Thus, the complete list of subgroups of S_3 is as follows: 1, \langle (1\ 2) \rangle, \langle (1\ 3) \rangle, \langle (2\ 3) \rangle, \langle (1\ 2\ 3) \rangle, and S_3.

Finally, the subgroup lattice of S_3 is as follows.

Find all normal subgroups of Dih(8) and the isomorphism type of each corresponding quotient

Find all normal subgroups of D_8 and for each of these find the isomorphism type of its corresponding quotient. (Use the subgroup lattice.)


The subgroup lattice of D_8 is as follows.

Lemma 1: For n > 2, \langle sr^k \rangle is not normal in D_{2n}. Proof: We have \langle sr^k \rangle = \{ 1, sr^k \}. Now r \langle sr^k \rangle = \{ r, sr^{k-1} \} and \langle sr^k \rangle r = \{ r, sr^{k+1} \}. Since n > 2, sr^{k-1} \neq sr^{k+1}, hence these sets are not equal and \langle sr^k \rangle is not normal. \square

Lemma 2: For n \geq 1, \langle r \rangle is normal in D_{2n}. Proof: Recall that D_{2n} = \langle r,s \rangle. Now srs^{-1} = r^{-1}s^2 = r^{-1} \in \langle r \rangle and rrr^{-1} = r \in \langle r \rangle. By a previous exercise, \langle r \rangle is normal in D_{2n}. \square.

Lemma 3: For n \geq 1, \langle s,r^2 \rangle is normal in D_{2n}. Proof: Recall that D_{2n} = \langle r,s \rangle. Now rsr^{-1} = sr^{-2} \in \langle s,r^2 \rangle, sss^{-1} = s \in \langle s,r^2 \rangle, rr^2r^{-1} = r^2 \in \langle s,r^2 \rangle, and sr^2s^{-1} = s^2r^{-2} \in \langle s,r^2 \rangle, so that by a previous exercise, \langle s,r^2 \rangle is normal in D_{2n}. \square

Lemma 4: For n \geq 1, n even, \langle sr, r^2 \rangle is normal in D_{2n}. Proof: Recall that D_{2n} = \langle r,s \rangle. Now r(sr)r^{-1} = sr^{-1} = sr(r^{n-2} = (sr)(r^2)^k \in \langle sr, r^2 \rangle for some k, ssrs^{-1} = sr^{-1} \in \langle sr, r^2 \rangle, rr^2r^{-1} = r^2 \in \langle sr,r^2 \rangle, and sr^2s^{-1} = r^{-2} \in \langle sr, r^2 \rangle, so that by a previous exercise, \langle sr, r^2 \rangle is normal in D_{2n} if n is even. \square

H H is normal/not normal because… D_8/H Because…
D_8 Trivially normal 1
\langle s,r^2 \rangle Normal by Lemma 3. Z_2 D_8/\langle s,r^2 \rangle has only two elements.
\langle r \rangle Normal by Lemma 2. Z_2 D_8/\langle r \rangle has only two elements.
\langle rs,r^2 \rangle Normal by Lemma 4. Z_2 D_8/\langle rs,r^2 \rangle has only two elements.
\langle s \rangle Not, by Lemma 1.
\langle r^2s \rangle Not, by Lemma 1.
\langle r^2 \rangle Normal because r^2 \in Z(D_8) by a previous exercise. V_4 D_8/\langle r^2 \rangle has order 4, but every nonidentity element has order 2.
\langle rs \rangle Not, by Lemma 1.
\langle r^3s \rangle Not, by Lemma 1.
1 Trivially normal D_8