## Tag Archives: cyclic group

### Exhibit some finite fields

Let $g(x) = x^2+x-1$ and $h(x) = x^3-x+1$. Construct finite fields of order 4, 8, 9, and 27 elements. For the fields with 4 and 9 elements, give the multiplication tables and show that the nonzero elements form a cyclic group.

Note that to construct fields of the given sizes, it suffices to show that $g(x)$ and $h(x)$ are each irreducible over the fields $\mathbb{F}_2$ and $\mathbb{F}_3$. (More precisely, that these polynomials are irreducible when we inject the coefficients into the respective fields.)

It is easy to see that neither $g$ nor $h$ has a root in either $\mathbb{F}_2$ or in $\mathbb{F}_3$, so these are indeed irreducible. Then $\mathbb{F}_2[x]/(g(x))$, $\mathbb{F}_2[x]/(h(x))$, $\mathbb{F}_3[x]/(g(x))$, and $\mathbb{F}_3[x]/(h(x))$ are fields of order 4, 8, 9, and 27 by Proposition 11 in D&F, and specifically are isomorphic as fields to the extension of $\mathbb{F}_2$ or $\mathbb{F}_3$ by adjoining a root of $g$ or $h$.

The multiplication table of $\mathbb{F}_2(\alpha) \cong \mathbb{F}_2[x]/(g(x))$ is as follows.

 $\begin{array}{c|cccc} & 0 & 1 & \alpha & \alpha+1 \\ \hline 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & \alpha & \alpha+1 \\ \alpha & 0 & \alpha & \alpha+1 & 1 \\ \alpha+1 & 0 & \alpha+1 & 1 & \alpha \end{array}$

Evidently, $(\mathbb{F}_2(\alpha))^\times$ is generated by $\alpha$.

The multiplication table of $\mathbb{F}_3(\alpha) \cong \mathbb{F}_3[x]/(g(x))$ is as follows.

 $\begin{array}{c|ccccccccc} & 0 & 1 & 2 & \alpha & \alpha+1 & \alpha+2 & 2\alpha & 2\alpha+1 & 2\alpha+2 \\ \hline 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 2 & \alpha & \alpha+1 & \alpha+2 & 2\alpha & 2\alpha+1 & 2\alpha+2 \\ 2 & 0 & 2 & 1 & 2\alpha & 2\alpha+2 & 2\alpha+1 & \alpha & \alpha+2 & \alpha+1 \\ \alpha & 0 & \alpha & 2\alpha & 2\alpha+1 & 1 & \alpha+1 & \alpha+2 & 2\alpha+2 & 2 \\ \alpha+1 & 0 & \alpha+1 & 2\alpha+2 & 1 & \alpha+2 & 2\alpha & 2 & \alpha & 2\alpha+1 \\ \alpha+2 & 0 & \alpha+2 & 2\alpha+1 & \alpha+1 & 2\alpha & 2 & 2\alpha+2 & 1 & \alpha \\ 2\alpha & 0 & 2\alpha & \alpha & \alpha+2 & 2 & 2\alpha+2 & 2\alpha+1 & \alpha+1 & 1 \\ 2\alpha+1 & 0 & 2\alpha+1 & \alpha+2 & 2\alpha+2 & \alpha & 1 & \alpha+1 & 2 & 2\alpha \\ 2\alpha+2 & 0 & 2\alpha+2 & \alpha+1 & 2 & 2\alpha+1 & \alpha & 1 & 2\alpha & \alpha+2 \end{array}$

Evidently, $(\mathbb{F}_3(\alpha))^\times$ is generated by $\alpha$.

### Compute Hom(ZZ/(m), ZZ/(n)) over ZZ

Compute $\mathsf{Hom}_\mathbb{Z}(\mathbb{Z}/(m), \mathbb{Z}/(n))$.

We claim that $\mathsf{Hom}_\mathbb{Z}(\mathbb{Z}/(m), \mathbb{Z}/(n)) \cong_\mathbb{Z} \mathbb{Z}/(m,n)$. (We use $\cong_R$ to mean “isomorphic as $R$-modules”.)

By this previous exercise, we have $\mathsf{Hom}_\mathbb{Z}(\mathbb{Z}/(m), \mathbb{Z}/(n)) \cong \mathsf{Ann}_{\mathbb{Z}/(n)}((m))$.

Let $d$ be a greatest common divisor of $m$ and $n$, and write $m = ad$ and $n = bd$ where $a$ and $b$ are relatively prime. Note that $m \cdot \overline{b} = a \cdot (d \cdot \overline{b})$ $= a \cdot (\overline{bd})$ $= a \cdot \overline{0} = \overline{0}$. So $\overline{b} \in \mathsf{Ann}_{\mathbb{Z}/(n)}((m))$. Now define $\varphi : \mathbb{Z} \rightarrow \mathsf{Ann}_{\mathbb{Z}/(n)}((m))$ by $k \mapsto k \cdot \overline{b}$; certainly $\varphi$ is a $\mathbb{Z}$-module homomorphism. We claim that $\varphi$ is surjective. To see this, suppose $\overline{t} \in \mathsf{Ann}_{\mathbb{Z}/(n)}((m))$. In particular, $m \cdot \overline{t} = \overline{mt} = \overline{0}$, so that $n|tm$. Then $b|at$, and since $a$ and $b$ are relatively prime, by Euclid’s lemma, we have $b|t$. So $t = bs$ for some $s$. Thus $\overline{t} = \varphi(s)$, and $\varphi$ is surjective. Next we claim that $\mathsf{ker}\ \varphi = (m,n) = (d)$. Certainly $\varphi(d) = d \cdot \overline{b} = \overline{0}$, so that $(d) \subseteq \mathsf{ker}\ \varphi$. Now suppose $s \in \mathsf{ker}\ \varphi$, so that $\varphi(s) = s \cdot \overline{b} = \overline{sb} = \overline{0}$. Then $n|sb$, so that $d|s$, and we have $s \in (d)$ as desired. By the First Isomorphism Theorem, we have $\mathsf{Ann}_{\mathbb{Z}/(n)}((m)) \cong \mathbb{Z}/(m,n)$.

Thus we have $\mathsf{Hom}_{\mathbb{Z}}(\mathbb{Z}/(m), \mathbb{Z}/(n)) \cong_R \mathbb{Z}/(m,n)$.

### Find a generating set for the augmentation ideal of a group ring

Let $R$ be a commutative ring with $1 \neq 0$ and $G$ a finite group. Prove that the augmentation ideal in the group ring $R[G]$ is generated by $\{ g-1 \ |\ g \in G \}$. Prove that if $G = \langle \sigma \rangle$ is cyclic then the augmentation ideal is generated by $\sigma - 1$.

Recall that the augmentation ideal of $R[G]$ is the kernel of the ring homomorphism $R[G] \rightarrow R$ given by $\sum r_ig_i \mapsto \sum r_i$; that is, it consists of all elements in $R[G]$ whose coefficients sum to 0 in $R$.

Let $S = \{ g-1 \ |\ g \in G \}$, and let $A$ denote the augmentation ideal of $R[G]$. First, note that $g-1 \in A$ for each $g \in G$, so that $(g-1) \subseteq A$. Thus $(S) \subseteq A$. Now let $\alpha = \sum_{g_i \in G} r_ig_i \in A$; then we have $\sum r_i = 0$. Consider the following.

 $\sum_{g_i \in G} r_i(g_i - 1)$ = $\sum_{g_i \in G} r_ig_i - r_i$ = $= \left( \sum_{g_i \in G} r_ig_i \right) - (\sum_{g_i \in G} r_i)$ = $\alpha - 0$ = $\alpha$

Thus $A \subseteq (S)$, and we have $A = (S)$.

Suppose further that $G = \langle \sigma \rangle$ is cyclic. We still have $(\sigma - 1) \subseteq A$. Now note that $(\sigma - 1)\left( \sum_{t=0}^k \sigma^t \right) = \sigma^{k+1} - 1$, so that $\sigma^k - 1 \in (\sigma - 1)$ for all $k$. Thus $A = (\sigma - 1)$.

### Characterization of free abelian groups of finite rank

Let $S$ be a set. The group presented by $(S, R)$ where $R = \{ [s,t] \ |\ s,t \in S \}$ is called the free abelian group on $S$. Denote it by $A(S)$. Prove that $A(S)$ has the following universal property: If $G$ is any abelian group and $\varphi S \rightarrow G$ is a set function, then there is a unique group homomorphism $\Phi : A(S) \rightarrow G$ such that $\Phi|_S = \varphi$. Deduce that if $A$ is a free abelian group on a set of (finite) cardinality $n$, then $A \cong \mathbb{Z}^n$.

Recall from the universal property of free groups that there is a unique group homomorphism $\Psi : F(S) \rightarrow G$ such that $\Psi|_S = \varphi$. Moreover, clearly $R \leq \mathsf{ker}\ \Psi$. Then there is a unique group homomorphism $\Psi : A(S) \rightarrow G$ such that $\Psi = \Phi \circ \pi$. Identifying $s$ in $S$ with $sR$ in $A(S)$, we have $\Phi|_S = \varphi$. This gives existence.

To see uniqueness, note that any homomorphism $\Phi : A(S) \rightarrow G$ extending $\varphi$ is uniquely determined by $\Psi$, which is uniquely determined by $\varphi$.

Now let $n \in \mathbb{N}$ and let $S = \{1, \ldots, n \}$. If $n = 0$, then $A(S)$ is a quotient of $F(S) = 1$, so that in fact $A(S) = 1$. Suppose $n > 0$, and for each $1 \leq k \leq n$, define $\varphi(k)$ to be the $n$-tuple with 1 in the $k$th coordinate and 0 in all other coordinates. By the universal property, there exists a unique group homomorphism $\Phi : A(S) \rightarrow \mathbb{Z}^n$ extending $\varphi$; $\Phi$ is surjective since $\Phi[S]$ generates $\mathbb{Z}^n$. Now if $w = s_1^{e_1} \cdots s_n^{e_n} \in \mathsf{ker}\ \Phi$, note that since $A(S)$ is abelian, we can collect terms so that $s_i = i$. Now $\Phi(w) = (e_1, \ldots, e_n) = (0, \ldots, 0)$, and in fact $w = 1$. Thus $\Phi$ is injective, and we have $A(S) \cong \mathbb{Z}^n$.

### The center of a subgroup of index 2 in a 2-group with cyclic center has rank at most 2

Use the preceding exercise to prove that if $P$ is a 2-group which has a cyclic center and $M$ is a subgroup of index 2 in $P$, then $Z(M)$ has rank at most 2.

Note that $M \leq P$ is normal and that $Z(M) \leq M$ is characteristic; thus $Z(M)$ is normal in $P$. Moreover, letting $Z(M)_2 = \{ z \in Z(M) \ |\ z^2 = 1 \}$, we know that $Z(M)_2 \leq Z(M)$ is characteristic, so that $Z(M)_2 \leq P$ is normal. Moreover, $Z(M)_2$ is elementary abelian by construction. Note that, by §5.2 #7, $Z(M)$ and $Z(M)_2$ have the same rank.

If $w \in P$, then conjugation by $w$ is an automorphism of $Z(M)_2$. If we choose $w \notin M$, and let $z$ be fixed by $w$– that is, $wz = zw$, then we have $z \in C_P(M)$ (since $z \in Z(M)$) and $z \in C_P(\langle w \rangle)$. Since $P = M \langle w \rangle$, in fact $w \in Z(P)$. Thus the subgroup of $Z(M)_2$ which is fixed under conjugation by $w$ is contained in $Z(P)$, and thus has rank 1.

By this previous exercise, $Z(M)_2$ has rank at most 2, so that $Z(M)$ also has rank at most 2.

### If the first two lower central quotients of a group are cyclic, then the second derived subgroup is trivial

Show that if $G^\prime/G^{\prime\prime}$ and $G^{\prime\prime} / G^{\prime\prime\prime}$ are both cyclic then $G^{\prime\prime} = 1$.

We begin by proving a lemma.

Lemma: Let $G$ be a group and let $N \leq G$ be normal. Then $(G/N)^\prime = G^\prime N /N$. Proof: $(\subseteq)$ If $[xN,yN] \in G/N$ is a commutator, then $[xN,yN] = [x,y]N \in G^\prime N/N$. $(\supseteq)$ $G^\prime \leq G^\prime N$, so that $G/G^\prime N \cong (G/N)/(G^\prime N/N)$ is abelian. By Theorem 5.7 in the text, $(G/N)^\prime \leq G^\prime N/N$. $\square$

Now for the main result.

Note that $G^{\prime\prime\prime} \leq G$ is characteristic, hence normal, so that $G^{\prime\prime}/G^{\prime\prime\prime} \leq G/G^{\prime\prime\prime}$ is a cyclic, hence abelian, normal subgroup.

By this previous exercise, $(G/G^{\prime\prime\prime})/(G^{\prime\prime}/G^{\prime\prime\prime}) \cong G/G^{\prime\prime}$ (via the Third Isomorphism Theorem) acts on $G^{\prime\prime}/G^{\prime\prime\prime}$ by conjugation on the left as follows: $(gG^{\prime\prime}) \cdot (a G^{\prime\prime\prime}) = (gag^{-1}) G^{\prime\prime\prime}$. Now conjugation is an automorphism of $G^{\prime\prime}/G^{\prime\prime\prime}$, and since this quotient is cyclic, its automorphism group is abelian. Thus, for all $g,h \in G$ and $a \in G^{\prime\prime}$, we have $(gh)a(gh)^{-1} G^{\prime\prime\prime} = (hg)a(hg)^{-1} G^{\prime\prime\prime}$. Via some arithmetic, we see that $[g,h] a [g,h]^{-1} a^{-1} \in G^{\prime\prime\prime}$; in particular, each element of $G^{\prime\prime}/G^{\prime\prime\prime}$ commutes with each generator of $G^{\prime}/G^{\prime\prime\prime}$, so that $G^{\prime\prime}/G^{\prime\prime\prime} \leq Z(G^{\prime}/G^{\prime\prime\prime})$.

Using the Third Isomorphism Theorem, $(G^\prime/G^{\prime\prime\prime})/Z(G^\prime/G^{\prime\prime\prime}) \cong ((G^\prime/G^{\prime\prime\prime})/(G^{\prime\prime}/G^{\prime\prime\prime}))/(Z(G^\prime/G^{\prime\prime\prime})/(G^{\prime\prime}/G^{\prime\prime\prime}))$ $\cong (G^\prime/G^{\prime\prime})/(Z(G^\prime/G^{\prime\prime\prime})/(G^{\prime\prime}/G^{\prime\prime\prime}))$ is cyclic; hence $G^\prime/G^{\prime\prime\prime}$ is abelian, and using the lemma we have $1 = (G^\prime/G^{\prime\prime\prime})^\prime = G^{\prime\prime}/G^{\prime\prime\prime}$. Thus $G^{\prime\prime} = G^{\prime\prime\prime}$.

### An equivalent characterization of finite cyclic groups

Prove the following proposition by invoking the Fundamental Theorem of Finite Abelian Groups.

Proposition: Let $G$ be a finite abelian group such that, for all positive integers $n$ dividing $|G|$, $G$ contains at most $n$ elements satisfying $x^n = 1$. Prove that $G$ is cyclic.

Proof: By FTFGAG, we have $G \cong Z_{n_1} \times \cdots \times Z_{n_t}$ for some integers $n_i$ such that $n_{i+1}|n_i$ for all $i$. Suppose $t \geq 2$, and let $p$ be a prime divisor of $n_2$. Now $p$ also divides $n_1$. Now $Z_{n_2}$ contains an element $x$ of order $p$, and also $Z_{n_1}$ contains an element $y$ of order $p$. Now every element $z \in \langle y \rangle \times \langle x \rangle$ satisfies $z^p = 1$, and there are $p^2$ such elements, a contradiction. Thus $t = 1$, hence $G = Z_{n_1}$ is cyclic.

### Some properties of nonabelian p groups of order p³

Prove that if $p$ is a prime and $P$ a nonabelian group of order $p^3$, then $|Z(P)| = p$ and $P/Z(P) \cong Z_p \times Z_p$.

By Lagrange, there are 4 possibilities for $|Z(P)|$: $1$, $p$, $p^2$, and $p^3$. However, we know that $Z(P)$ is nontrivial, and if $|Z(P)| = p^3$, then $P$ is abelian.

If $|Z(P)| = p^2$, then $P/Z(P) \cong Z_p$ is cyclic, and we have $P$ abelian, a contradiction.

If $|Z(P)| = p$, there are two possibilities for the isomorphism type of $P/Z(P)$: $Z_{p^2}$ and $Z_p \times Z_p$. In the first case, $P$ is abelian. Thus $P/Z(P) \cong Z_p \times Z_p$.

### Equivalent characterizations of nilpotent and cyclic groups

If $G$ is finite, prove that (1) $G$ is nilpotent if and only if it has a normal subgroup of each order dividing $|G|$, and (2) is cyclic if and only if it has a unique subgroup of each order dividing $|G|$.

We begin with some lemmas.

Lemma 1: Let $G$ be a group, and let $H_1, \ldots, H_k$ be characteristic subgroups of $G$. Then $H = H_1\cdots H_k$ is characteristic in $G$. Proof: $H$ is a subgroup because each $H_i$ is normal. If $\alpha$ is an automorphism of $G$, then $\alpha[H] = \alpha[H_1] \cdots \alpha[H_k]$ $= H_1 \cdots H_k = H$. $\square$

Lemma 2: Let $G$ be a finite group. If $P \leq G$ is a normal Sylow subgroup, then $P$ is characteristic. Proof: $P$ is the unique subgroup of order $|P|$. $\square$

Now to the main result; first we prove (1).

$(\Rightarrow)$ We proceed by induction on the breadth (number of prime divisors with multiplicity) of $G$.

For the base case, if $G$ is a finite nilpotent group of breadth 1, then $G \cong Z_p$ is an abelian simple group. Thus $G$ has a normal subgroup of order 1 and $p$, the only divisors of $|G|$.

For the inductive step, suppose that every finite nilpotent group of breadth $k \geq 1$ has a normal subgroup of each order dividing $|G|$, and let $G$ be a finite nilpotent group of breadth $k+1$. Now $Z(G)$ is nontrivial, so there exists an element $x \in Z(G)$ of prime order, and $\langle x \rangle G$ is normal. We showed in a lemma to the previous exercise that $G/\langle x \rangle$ is nilpotent, and has breadth $k$. By the inductive hypothesis, $G/\langle x \rangle$ has a normal subgroup of each order $d$ dividing $|G/\langle x \rangle|$. By the Lattice Isomorphism Theorem, a subgroup $\overline{H} \leq G/\langle x \rangle$ is normal if and only if $\overline{H} = H/\langle x \rangle$ for some normal subgroup $H \leq G$. Thus we see that $G$ has a normal subgroup of order $d$ for all $d$ dividing $|G|$ such that $p|d$. Now let $d$ divide $|G|$, with $p$ not dividing $d$, and let $H \leq G$ be a normal subgroup of order $pd$. Since $H \leq G$, $H$ is a finite nilpotent group. By Theorem 3, $H$ is the direct product of its Sylow subgroups. Write $H = P \times Q$, where $P$ is the Sylow $p$ subgroup of $H$ and $Q$ the direct product of the remaining Sylow subgroups. By Lemma 2, each factor of $Q$ is characteristic in $H$, so that $Q$ is characteristic in $H$. Note that $|Q| = d$, thus $Q \leq G$ is normal and has order $d$.

$(\Leftarrow)$ Suppose $G$ has a normal subgroup of each order $d$ dividing $|G|$. In particular, $G$ has a normal (thus unique) Sylow $p$-subgroup for each prime $p$. By Theorem 6.3, $G$ is nilpotent.

Now we prove (2).

The $(\Rightarrow)$ direction was proved in Theorem 2.7.

$(\Leftarrow)$ Suppose $G$ is finite and has a unique subgroup of order $d$ for each $d$ properly dividing $|G|$. In particular, all the Sylow subgroups of $G$ are unique, hence normal; say there are $k$ of these. Suppose some Sylow $p$-subgroup $P$ of $G$ is not cyclic; then by this previous exercise, $P$ has a subgroup congruent to $Z_p \times Z_p$, and thus $G$ has distinct subgroups of order $p$, a contradiction. Thus each Sylow subgroup of $G$ is cyclic; say $P_i = \langle x_i \rangle$. Then $|x_1 \cdots x_k| = |G|$, so that $G = \langle x_1 \cdots x_k \rangle$ is cyclic.

### Construct all the semidirect products of Cyc(2ⁿ) by Cyc(2)

Show that for any $n \geq 3$ there are exactly 4 distinct homomorphisms from $Z_2$ into $\mathsf{Aut}(Z_{2^n})$. Prove that the resulting semidirect products are nonisomorphic groups of order $2^{n+1}$. (These four groups, with the cyclic group and the generalized quaternion group, are all the groups of order $2^{n+1}$ which have a cyclic subgroup of index 2.)

We begin with two lemmas.

Lemma 1: Let $k \geq 3$ be an integer. Suppose $m$ and $n$ are integers such that, for all integers $a$ and $b$, $m+a(1+2^{k-1})^n \equiv a+m(1+2^{k-1})^b$ mod $2^k$. Then $m \equiv n \equiv 0$ mod 2. Proof: Note that $(1+2^{k-1})^2 \equiv 1$ mod $2^k$. Suppose $n \equiv 1$ mod 2. Let $b = 2$. Then we have $a2^{k-1} \equiv 0$ mod $2^k$, a contradiction when $a$ is odd. Thus $n \equiv 0$ mod $2^k$. Now choosing $b$ odd, we have $m2^{k-1} \equiv 0$ mod $2^k$; thus $m \equiv 0$ mod 2. $\square$

Lemma 2: Let $k \geq 3$ be an integer. Suppose $m$ and $n$ are integers such that, for all integers $a$ and $b$, $m+a(-1+2^{k-1})^n \equiv a+m(-1+2^{k-1})^b$ mod $2^k$. Then $m \equiv 0$ mod $2^{k-1}$ and $n \equiv 0$ mod 2. Proof: Note that $(-1+2^{k-1})^2 \equiv 1$ mod $2^k$. Suppose $n \equiv 1$ mod 2. Let $b = 2$. Then we have $0 \equiv 2a(1-2^{k-2})$ mod $2^k$. Since $1-2^{k-2}$ is odd, we have $2a \equiv 0$ mod $2^k$ for all $a$, which is absurd. Thus $n \equiv 0$ mod 2. Now let $b$ be odd; we have $2m(1-2^{k-2}) \equiv 0$ mod $2^k$; again since $1-2^{k-2}$ is odd we have $2m \equiv 0$ mod $2^k$, so that $m \equiv 0$ mod $2^{k-1}$. $\square$

We found that there are 4 distinct homomorphisms $Z_2 \rightarrow \mathsf{Aut}(Z_{2^n})$ in two lemmas to the previous exercise. Moreover, writing $Z_2 = \langle x \rangle$ and $Z_{2^n} = \langle y \rangle$, these are given as follows. $\varphi_1(x)(y) = y$, $\varphi_2(x)(y) = y^{-1}$, $\varphi_3(x)(y) = y^{1+2^{n-1}}$, and $\varphi_4(x)(y) = y^{-1+2^{n-1}}$.

Let $G_i = Z_{2^n} \rtimes_{\varphi_i} Z_2$. We will now show that these $G_i$ are distinct.

Note that $\varphi_1$ is trivial, so that $G_1 \cong Z_{2^n} \times Z_2$ is abelian. However, the remaining groups are not abelian, so $G_1$ is distinct from the remaining three.

Note that $G_2 \cong D_{2^{n+1}} = \langle r,s \rangle$. By this previous exercise, $Z(D_{2^{n+1}}) = \langle r^{2^{n-1}} \rangle \cong Z_2$.

Now $G_3$ is generated by $x$ and $y$, and we have $xyx^{-1} = \varphi_3(x)(y) = y^{1+2^{n-1}}$. Thus this group has the presentation $\langle x,y \ |\ x^2 = y^{2^n} = 1, xy = y^{1+2^{n-1}}x \rangle$. Similarly, $G_4$ has the presentation $\langle x,y \ |\ x^2 = y^{2^n} = 1, xy = y^{-1+2^{n-1}}x \rangle$.

We now compute the centers of $G_3$ and $G_4$.

Every element of $G_3$ can be written as $y^ix^j$ where $0 < i \leq 2^n$ and $0 < j \leq 2$, and there are $2^{n+1}$ such expressions. Since we know by other means that $|G_3| = 2^{n+1}$, every element of $G_3$ can be written in this form uniquely. Let $s = y^ax^b \in Z(G_3)$ and let $y^ix^j \in G_3$. Then $y^ax^by^ix^j = y^{a+i(1+2^{n-1})^b}x^{b+j}$ is equal to $y^ix^jy^ax^b = y^{i+a(1+2^{n-1})^j}x^{b+j}$. Comparing exponents and using Lemma 1, we have $a \equiv b \equiv 0$ mod 2. Thus $s \in \langle y^2 \rangle$. Now note that $y^2y^ix^j = y^{i+2}x^j$ and $y^ixy^2 = y^iy^{2(1+2^{n-1})}x = y^{i+2}x$, so that $y^2 \in Z(G_3)$; thus $Z(G_3) = \langle y^2 \rangle \cong Z_{2^{n-1}}$.

Performing the same analysis on $G_4$ reveals that $Z(G_4) = \langle y^{2^{n-1}} \rangle \cong Z_2$.

Thus $G_3$ is distinct from $G_2$ and $G_4$.

We know that $G_2 \cong D_{2^{n+1}}$ has $2^n + 1$ elements of order 2. Suppose $s = y^ax^b \in G_4$ has order 2; then $(y^ax^b)^2 = y^{a(1+(-1+2^{n-1})^b)} = 1$. If $b = 0$, then $a = 0$, so that $s = 1$. If $b = 1$, then $a2^{n-1} \equiv 0$ mod $2^n$, so that $a \equiv 0$ mod 2. There are $2^{n-1}$ such numbers between 0 and $2^n$. Thus $G_4$ has $2^{n-1}$ elements of order 2, and $G_2 \not\cong G_4$.