## 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 $D_8$