Sample Solution

MMT-003 Solved Assignment 2025

The statement "If a finite group $G$$G$GG$G$ acts on a finite set $S$$S$SS$S$, then $G_{s_1}=G_{s_2}$$G_{s_1}=G_{s_2}$G_(s_(1))=G_(s_(2))G_{s_1}=G_{s_2}$G_{s_1}=G_{s_2}$ for all $s_1,s_2\in S$$s_1,s_2\in S$s_(1),s_(2)in Ss_1, s_2 \in S$s_1,s_2\in S$" is false. Here, $G_{s_1}$$G_{s_1}$G_(s_(1))G_{s_1}$G_{s_1}$ and $G_{s_2}$$G_{s_2}$G_(s_(2))G_{s_2}$G_{s_2}$ denote the stabilizer subgroups of $s_1$$s_1$s_(1)s_1$s_1$ and $s_2$$s_2$s_(2)s_2$s_2$ in $S$$S$SS$S$, respectively. The stabilizer subgroup $G_s$$G_s$G_(s)G_s$G_s$ of an element $s\in S$$s\in S$s in Ss \in S$s\in S$ is defined as the set of all elements in $G$$G$GG$G$ that fix $s$$s$ss$s$, i.e., $G_s=\{g\in G\mid g\cdot s=s\}$$G_s=\{g\in G\mid g\cdot s=s\}$G_(s)={g in G∣g*s=s}G_s = \{g \in G \mid g \cdot s = s\}$G_s=\{g\in G\mid g\cdot s=s\}$.

The statement claims that for any two elements $s_1$$s_1$s_(1)s_1$s_1$ and $s_2$$s_2$s_(2)s_2$s_2$ in the set $S$$S$SS$S$, their stabilizer subgroups in $G$$G$GG$G$ are equal. This would mean that every element of $G$$G$GG$G$ that fixes $s_1$$s_1$s_(1)s_1$s_1$ also fixes $s_2$$s_2$s_(2)s_2$s_2$ and vice versa. However, this is not necessarily true for all actions of $G$$G$GG$G$ on $S$$S$SS$S$. The equality of stabilizers for all elements in $S$$S$SS$S$ would imply a very rigid action, essentially saying that either the entire group fixes every element of $S$$S$SS$S$ or that the action is trivial (where $G$$G$GG$G$ acts on $S$$S$SS$S$ in such a way that every element of $G$$G$GG$G$ fixes every element of $S$$S$SS$S$).

Consider the group $G=\{e,g\}$$G=\{e,g\}$G={e,g}G = \{e, g\}$G=\{e,g\}$ where $e$$e$ee$e$ is the identity element and $g$$g$gg$g$ is an element of order 2 (meaning $g^2=e$$g^2=e$g^(2)=eg^2 = e$g^2=e$), acting on the set $S=\{s_1,s_2\}$$S=\{s_1,s_2\}$S={s_(1),s_(2)}S = \{s_1, s_2\}$S=\{s_1,s_2\}$ with two elements. Define the action of $G$$G$GG$G$ on $S$$S$SS$S$ as follows:

For this action, we have:

In this specific example, $G_{s_1}$$G_{s_1}$G_(s_(1))G_{s_1}$G_{s_1}$ and $G_{s_2}$$G_{s_2}$G_(s_(2))G_{s_2}$G_{s_2}$ happen to be equal because the non-identity element of $G$$G$GG$G$ does not fix either of the elements of $S$$S$SS$S$ but rather swaps them. However, the action of $g$$g$gg$g$ clearly differentiates $s_1$$s_1$s_(1)s_1$s_1$ and $s_2$$s_2$s_(2)s_2$s_2$, showing that the nature of the action can lead to different behaviors for different elements of $S$$S$SS$S$, and it’s easy to construct examples where $G_{s_1}\ne G_{s_2}$$G_{s_1}\ne G_{s_2}$G_(s_(1))!=G_(s_(2))G_{s_1} \neq G_{s_2}$G_{s_1}\ne G_{s_2}$ by adjusting the action.

For a more direct counterexample to the statement, consider a group $G$$G$GG$G$ acting on a set $S=\{s_1,s_2,s_3\}$$S=\{s_1,s_2,s_3\}$S={s_(1),s_(2),s_(3)}S = \{s_1, s_2, s_3\}$S=\{s_1,s_2,s_3\}$ where $G$$G$GG$G$ fixes $s_1$$s_1$s_(1)s_1$s_1$ and $s_2$$s_2$s_(2)s_2$s_2$ but not $s_3$$s_3$s_(3)s_3$s_3$, or any scenario where the action is not uniform across all elements of $S$$S$SS$S$. This shows that the original statement is false.

(b) There are exactly 8 elements of order 3 in $S_4$$S_4$S_(4)S_4$S_4$.

To determine the truth of the statement "There are exactly 8 elements of order 3 in $S_4$$S_4$S_(4)S_4$S_4$," we need to understand the structure of the symmetric group $S_4$$S_4$S_(4)S_4$S_4$ and the concept of an element’s order.

Elements of order 3 in $S_4$$S_4$S_(4)S_4$S_4$ are those permutations that cycle three elements while leaving the fourth unchanged. These can be represented as 3-cycles, such as $(123)$$(123)$(123)(123)$(123)$, which means that 1 goes to 2, 2 goes to 3, and 3 goes back to 1, with the fourth element (in this case, 4) remaining fixed.

To count the elements of order 3, we look for all possible 3-cycles within $S_4$$S_4$S_(4)S_4$S_4$. A 3-cycle can be chosen by selecting 3 elements out of 4 to participate in the cycle, and the order in which these three elements are arranged matters (since $(123)$$(123)$(123)(123)$(123)$ is different from $(132)$$(132)$(132)(132)$(132)$, for example).

The number of ways to choose 3 elements out of 4 is given by the combination formula $(\genfrac{}{}{0ex}{}{4}{3})$$(\genfrac{}{}{0ex}{}{4}{3})$((4)/(3))\binom{4}{3}$(\genfrac{}{}{0ex}{}{4}{3})$. For each selection of 3 elements, there are 2 ways to arrange them into a cycle (since for any three distinct elements $a$$a$aa$a$, $b$$b$bb$b$, and $c$$c$cc$c$, there are exactly two 3-cycles: $(abc)$$(abc)$(abc)(abc)$(abc)$ and $(acb)$$(acb)$(acb)(acb)$(acb)$).

Thus, the total number of 3-cycles (elements of order 3) in $S_4$$S_4$S_(4)S_4$S_4$ is:

Let’s calculate this value:

After calculating, we find that there are indeed exactly 8 elements of order 3 in $S_4$$S_4$S_(4)S_4$S_4$, which means the statement is true. Each of these elements is a permutation that cycles three of the four elements back to their original positions after three applications, which is the definition of having order 3.

(c) If $F=\mathbb{Q}(\sqrt[5]{2},\sqrt[3]{5})$$F=\mathbb{Q}(\sqrt[5]{2},\sqrt[3]{5})$F=Q(root(5)(2),root(3)(5))F=\mathbb{Q}(\sqrt[5]{2}, \sqrt[3]{5})$F=\mathbb{Q}(\sqrt[5]{2},\sqrt[3]{5})$, then $[F:\mathbb{Q}]=8$$[F:\mathbb{Q}]=8$[F:Q]=8[F: \mathbb{Q}]=8$[F:\mathbb{Q}]=8$.

To evaluate the statement "If $F=\mathbb{Q}(\sqrt[5]{2},\sqrt[3]{5})$$F=\mathbb{Q}(\sqrt[5]{2},\sqrt[3]{5})$F=Q(root(5)(2),root(3)(5))F=\mathbb{Q}(\sqrt[5]{2}, \sqrt[3]{5})$F=\mathbb{Q}(\sqrt[5]{2},\sqrt[3]{5})$, then $[F:\mathbb{Q}]=8$$[F:\mathbb{Q}]=8$[F:Q]=8[F: \mathbb{Q}]=8$[F:\mathbb{Q}]=8$," we need to understand the concept of field extensions and the degree of an extension.

Given that the degrees of the individual extensions are 5 and 3, respectively, and assuming no prior knowledge about a direct interaction that would reduce the combined degree, one might naively expect the degree of the combined extension to be $5\times 3=15$$5\times 3=15$5xx3=155 \times 3 = 15$5\times 3=15$. This is because the extensions are generated by roots of polynomials that are irreducible over $\mathbb{Q}$$\mathbb{Q}$Q\mathbb{Q}$\mathbb{Q}$ and do not immediately suggest a common field that would reduce the degree of the extension.

The statement "If $F=\mathbb{Q}(\sqrt[5]{2},\sqrt[3]{5})$$F=\mathbb{Q}(\sqrt[5]{2},\sqrt[3]{5})$F=Q(root(5)(2),root(3)(5))F=\mathbb{Q}(\sqrt[5]{2}, \sqrt[3]{5})$F=\mathbb{Q}(\sqrt[5]{2},\sqrt[3]{5})$, then $[F:\mathbb{Q}]=8$$[F:\mathbb{Q}]=8$[F:Q]=8[F: \mathbb{Q}]=8$[F:\mathbb{Q}]=8$" is false. Based on the analysis above, without specific interaction that would reduce the degree, the expected degree of the field extension $F$$F$FF$F$ over $\mathbb{Q}$$\mathbb{Q}$Q\mathbb{Q}$\mathbb{Q}$ would naively be calculated as $15$$15$1515$15$, not $8$$8$88$8$, assuming independent extensions generated by $\sqrt[5]{2}$$\sqrt[5]{2}$root(5)(2)\sqrt[5]{2}$\sqrt[5]{2}$ and $\sqrt[3]{5}$$\sqrt[3]{5}$root(3)(5)\sqrt[3]{5}$\sqrt[3]{5}$. However, the actual degree calculation requires detailed knowledge of how these extensions interact, and in general, the degree of such a combined extension is not simply the product of the degrees if there’s a nontrivial intersection. But in this specific case, without evidence of such an intersection that would lower the degree to $8$$8$88$8$, the initial analysis suggests the statement is incorrect.

(d) ${\mathbf{F}}_7(\sqrt{3})={\mathbf{F}}_7(\sqrt{5})$${\mathbf{F}}_7(\sqrt{3})={\mathbf{F}}_7(\sqrt{5})$F_(7)(sqrt3)=F_(7)(sqrt5)\mathbf{F}_7(\sqrt{3})=\mathbf{F}_7(\sqrt{5})${\mathbf{F}}_7(\sqrt{3})={\mathbf{F}}_7(\sqrt{5})$.

To evaluate the statement "${\mathbf{F}}_7(\sqrt{3})={\mathbf{F}}_7(\sqrt{5})$${\mathbf{F}}_7(\sqrt{3})={\mathbf{F}}_7(\sqrt{5})$F_(7)(sqrt3)=F_(7)(sqrt5)\mathbf{F}_7(\sqrt{3})=\mathbf{F}_7(\sqrt{5})${\mathbf{F}}_7(\sqrt{3})={\mathbf{F}}_7(\sqrt{5})$," we need to understand the structure of finite fields, specifically the field extensions of ${\mathbf{F}}_7$${\mathbf{F}}_7$F_(7)\mathbf{F}_7${\mathbf{F}}_7$, the finite field with 7 elements.

To determine whether ${\mathbf{F}}_7(\sqrt{3})={\mathbf{F}}_7(\sqrt{5})$${\mathbf{F}}_7(\sqrt{3})={\mathbf{F}}_7(\sqrt{5})$F_(7)(sqrt3)=F_(7)(sqrt5)\mathbf{F}_7(\sqrt{3})=\mathbf{F}_7(\sqrt{5})${\mathbf{F}}_7(\sqrt{3})={\mathbf{F}}_7(\sqrt{5})$, we need to check if $\sqrt{3}$$\sqrt{3}$sqrt3\sqrt{3}$\sqrt{3}$ and $\sqrt{5}$$\sqrt{5}$sqrt5\sqrt{5}$\sqrt{5}$ can be expressed in terms of each other within the field extensions of ${\mathbf{F}}_7$${\mathbf{F}}_7$F_(7)\mathbf{F}_7${\mathbf{F}}_7$.

Let’s perform explicit checks for the existence of $\sqrt{3}$$\sqrt{3}$sqrt3\sqrt{3}$\sqrt{3}$ and $\sqrt{5}$$\sqrt{5}$sqrt5\sqrt{5}$\sqrt{5}$ in ${\mathbf{F}}_7$${\mathbf{F}}_7$F_(7)\mathbf{F}_7${\mathbf{F}}_7$:

Let’s calculate these explicitly to see if such elements exist and to understand the relationship between ${\mathbf{F}}_7(\sqrt{3})$${\mathbf{F}}_7(\sqrt{3})$F_(7)(sqrt3)\mathbf{F}_7(\sqrt{3})${\mathbf{F}}_7(\sqrt{3})$ and ${\mathbf{F}}_7(\sqrt{5})$${\mathbf{F}}_7(\sqrt{5})$F_(7)(sqrt5)\mathbf{F}_7(\sqrt{5})${\mathbf{F}}_7(\sqrt{5})$.

After calculating, we find that there are no integer solutions for $x^2\equiv 3\mathrm{mod}7$$x^2\equiv 3\mathrm{mod}7$x^(2)-=3mod7x^2 \equiv 3 \mod 7$x^2\equiv 3\mathrm{mod}7$ and $x^2\equiv 5\mathrm{mod}7$$x^2\equiv 5\mathrm{mod}7$x^(2)-=5mod7x^2 \equiv 5 \mod 7$x^2\equiv 5\mathrm{mod}7$ within ${\mathbf{F}}_7$${\mathbf{F}}_7$F_(7)\mathbf{F}_7${\mathbf{F}}_7$. This means that neither $\sqrt{3}$$\sqrt{3}$sqrt3\sqrt{3}$\sqrt{3}$ nor $\sqrt{5}$$\sqrt{5}$sqrt5\sqrt{5}$\sqrt{5}$ can be directly represented by any element in ${\mathbf{F}}_7$${\mathbf{F}}_7$F_(7)\mathbf{F}_7${\mathbf{F}}_7$ through simple squaring and modular arithmetic.

However, the statement "${\mathbf{F}}_7(\sqrt{3})={\mathbf{F}}_7(\sqrt{5})$${\mathbf{F}}_7(\sqrt{3})={\mathbf{F}}_7(\sqrt{5})$F_(7)(sqrt3)=F_(7)(sqrt5)\mathbf{F}_7(\sqrt{3})=\mathbf{F}_7(\sqrt{5})${\mathbf{F}}_7(\sqrt{3})={\mathbf{F}}_7(\sqrt{5})$" refers to the equality of field extensions of ${\mathbf{F}}_7$${\mathbf{F}}_7$F_(7)\mathbf{F}_7${\mathbf{F}}_7$ that include $\sqrt{3}$$\sqrt{3}$sqrt3\sqrt{3}$\sqrt{3}$ and $\sqrt{5}$$\sqrt{5}$sqrt5\sqrt{5}$\sqrt{5}$, respectively. Given that neither $\sqrt{3}$$\sqrt{3}$sqrt3\sqrt{3}$\sqrt{3}$ nor $\sqrt{5}$$\sqrt{5}$sqrt5\sqrt{5}$\sqrt{5}$ has a direct representation in ${\mathbf{F}}_7$${\mathbf{F}}_7$F_(7)\mathbf{F}_7${\mathbf{F}}_7$, both $\sqrt{3}$$\sqrt{3}$sqrt3\sqrt{3}$\sqrt{3}$ and $\sqrt{5}$$\sqrt{5}$sqrt5\sqrt{5}$\sqrt{5}$ would necessitate extensions of ${\mathbf{F}}_7$${\mathbf{F}}_7$F_(7)\mathbf{F}_7${\mathbf{F}}_7$ to be included.

The absence of integer solutions for these equations in ${\mathbf{F}}_7$${\mathbf{F}}_7$F_(7)\mathbf{F}_7${\mathbf{F}}_7$ suggests that both $\sqrt{3}$$\sqrt{3}$sqrt3\sqrt{3}$\sqrt{3}$ and $\sqrt{5}$$\sqrt{5}$sqrt5\sqrt{5}$\sqrt{5}$ would lead to the creation of quadratic extensions of ${\mathbf{F}}_7$${\mathbf{F}}_7$F_(7)\mathbf{F}_7${\mathbf{F}}_7$, as both would be roots of irreducible quadratic polynomials over ${\mathbf{F}}_7$${\mathbf{F}}_7$F_(7)\mathbf{F}_7${\mathbf{F}}_7$. The key insight here is not the absence of solutions within ${\mathbf{F}}_7$${\mathbf{F}}_7$F_(7)\mathbf{F}_7${\mathbf{F}}_7$ but the nature of the extensions they generate.

Given the structure of finite fields, any extension of ${\mathbf{F}}_7$${\mathbf{F}}_7$F_(7)\mathbf{F}_7${\mathbf{F}}_7$ to include a square root not in ${\mathbf{F}}_7$${\mathbf{F}}_7$F_(7)\mathbf{F}_7${\mathbf{F}}_7$ would have to be a quadratic extension, implying that both ${\mathbf{F}}_7(\sqrt{3})$${\mathbf{F}}_7(\sqrt{3})$F_(7)(sqrt3)\mathbf{F}_7(\sqrt{3})${\mathbf{F}}_7(\sqrt{3})$ and ${\mathbf{F}}_7(\sqrt{5})$${\mathbf{F}}_7(\sqrt{5})$F_(7)(sqrt5)\mathbf{F}_7(\sqrt{5})${\mathbf{F}}_7(\sqrt{5})$ are quadratic extensions of ${\mathbf{F}}_7$${\mathbf{F}}_7$F_(7)\mathbf{F}_7${\mathbf{F}}_7$. Since all quadratic extensions of a given finite field have the same size (in this case, $49$$49$4949$49$ elements, or ${\mathbf{F}}_{7^2}$${\mathbf{F}}_{7^2}$F_(7^(2))\mathbf{F}_{7^2}${\mathbf{F}}_{7^2}$), and since there’s only one field of each possible order in the realm of finite fields, it follows that ${\mathbf{F}}_7(\sqrt{3})$${\mathbf{F}}_7(\sqrt{3})$F_(7)(sqrt3)\mathbf{F}_7(\sqrt{3})${\mathbf{F}}_7(\sqrt{3})$ and ${\mathbf{F}}_7(\sqrt{5})$${\mathbf{F}}_7(\sqrt{5})$F_(7)(sqrt5)\mathbf{F}_7(\sqrt{5})${\mathbf{F}}_7(\sqrt{5})$ must indeed be the same field, despite the initial intuition suggested by the absence of direct square roots in ${\mathbf{F}}_7$${\mathbf{F}}_7$F_(7)\mathbf{F}_7${\mathbf{F}}_7$.

Therefore, the statement "${\mathbf{F}}_7(\sqrt{3})={\mathbf{F}}_7(\sqrt{5})$${\mathbf{F}}_7(\sqrt{3})={\mathbf{F}}_7(\sqrt{5})$F_(7)(sqrt3)=F_(7)(sqrt5)\mathbf{F}_7(\sqrt{3})=\mathbf{F}_7(\sqrt{5})${\mathbf{F}}_7(\sqrt{3})={\mathbf{F}}_7(\sqrt{5})$" is true, as both $\sqrt{3}$$\sqrt{3}$sqrt3\sqrt{3}$\sqrt{3}$ and $\sqrt{5}$$\sqrt{5}$sqrt5\sqrt{5}$\sqrt{5}$ lead to the same quadratic extension of ${\mathbf{F}}_7$${\mathbf{F}}_7$F_(7)\mathbf{F}_7${\mathbf{F}}_7$, given the unique structure of finite fields and their extensions.

(e) For any $\alpha \in {\mathbb{F}}_{2^5}^{\ast },\alpha \ne 1,{\mathbb{F}}_{2^5}={\mathbb{F}}_2[\alpha ]$$\alpha \in {\mathbb{F}}_{2^5}^{\ast },\alpha \ne 1,{\mathbb{F}}_{2^5}={\mathbb{F}}_2[\alpha ]$alpha inF_(2^(5))^(**),alpha!=1,F_(2^(5))=F_(2)[alpha]\alpha \in \mathbb{F}_{2^5}^*, \alpha \neq 1, \mathbb{F}_{2^5}=\mathbb{F}_2[\alpha]$\alpha \in {\mathbb{F}}_{2^5}^{\ast },\alpha \ne 1,{\mathbb{F}}_{2^5}={\mathbb{F}}_2[\alpha ]$.

The statement "For any $\alpha \in {\mathbb{F}}_{2^5}^{\ast },\alpha \ne 1,{\mathbb{F}}_{2^5}={\mathbb{F}}_2[\alpha ]$$\alpha \in {\mathbb{F}}_{2^5}^{\ast },\alpha \ne 1,{\mathbb{F}}_{2^5}={\mathbb{F}}_2[\alpha ]$alpha inF_(2^(5))^(**),alpha!=1,F_(2^(5))=F_(2)[alpha]\alpha \in \mathbb{F}_{2^5}^*, \alpha \neq 1, \mathbb{F}_{2^5}=\mathbb{F}_2[\alpha]$\alpha \in {\mathbb{F}}_{2^5}^{\ast },\alpha \ne 1,{\mathbb{F}}_{2^5}={\mathbb{F}}_2[\alpha ]$" concerns the structure of finite fields, specifically the field with $2^5=32$$2^5=32$2^(5)=322^5 = 32$2^5=32$ elements, denoted ${\mathbb{F}}_{2^5}$${\mathbb{F}}_{2^5}$F_(2^(5))\mathbb{F}_{2^5}${\mathbb{F}}_{2^5}$, and the concept of field extensions.

The statement asserts that for any non-identity element $\alpha$$\alpha$alpha\alpha$\alpha$ of ${\mathbb{F}}_{2^5}^{\ast }$${\mathbb{F}}_{2^5}^{\ast }$F_(2^(5))^(**)\mathbb{F}_{2^5}^*${\mathbb{F}}_{2^5}^{\ast }$, the field generated by $\alpha$$\alpha$alpha\alpha$\alpha$ over ${\mathbb{F}}_2$${\mathbb{F}}_2$F_(2)\mathbb{F}_2${\mathbb{F}}_2$ is the entire field ${\mathbb{F}}_{2^5}$${\mathbb{F}}_{2^5}$F_(2^(5))\mathbb{F}_{2^5}${\mathbb{F}}_{2^5}$. This would mean that every non-identity element in ${\mathbb{F}}_{2^5}$${\mathbb{F}}_{2^5}$F_(2^(5))\mathbb{F}_{2^5}${\mathbb{F}}_{2^5}$ is a primitive element.

However, this is not true for every non-identity element of ${\mathbb{F}}_{2^5}$${\mathbb{F}}_{2^5}$F_(2^(5))\mathbb{F}_{2^5}${\mathbb{F}}_{2^5}$. While it is true that a primitive element $\alpha$$\alpha$alpha\alpha$\alpha$ (an element of order $31$$31$3131$31$) will generate the entire field ${\mathbb{F}}_{2^5}$${\mathbb{F}}_{2^5}$F_(2^(5))\mathbb{F}_{2^5}${\mathbb{F}}_{2^5}$ through its powers, not every non-identity element in ${\mathbb{F}}_{2^5}$${\mathbb{F}}_{2^5}$F_(2^(5))\mathbb{F}_{2^5}${\mathbb{F}}_{2^5}$ is a primitive element. Some elements will have smaller orders due to the structure of the cyclic group ${\mathbb{F}}_{2^5}^{\ast }$${\mathbb{F}}_{2^5}^{\ast }$F_(2^(5))^(**)\mathbb{F}_{2^5}^*${\mathbb{F}}_{2^5}^{\ast }$. For example, an element of order $3$$3$33$3$ would only cycle through $3$$3$33$3$ distinct non-zero elements (plus the identity element when raised to the power of $3$$3$33$3$) before repeating, and thus, it cannot generate the entire field.

To disprove the statement, we can consider the property of the elements’ orders in ${\mathbb{F}}_{2^5}^{\ast }$${\mathbb{F}}_{2^5}^{\ast }$F_(2^(5))^(**)\mathbb{F}_{2^5}^*${\mathbb{F}}_{2^5}^{\ast }$. The order of any element $\alpha$$\alpha$alpha\alpha$\alpha$ in a finite field must divide the order of the field’s multiplicative group, which is $31$$31$3131$31$ in this case (since ${\mathbb{F}}_{2^5}^{\ast }$${\mathbb{F}}_{2^5}^{\ast }$F_(2^(5))^(**)\mathbb{F}_{2^5}^*${\mathbb{F}}_{2^5}^{\ast }$ has $31$$31$3131$31$ elements). By Lagrange’s theorem, the possible orders of elements in ${\mathbb{F}}_{2^5}^{\ast }$${\mathbb{F}}_{2^5}^{\ast }$F_(2^(5))^(**)\mathbb{F}_{2^5}^*${\mathbb{F}}_{2^5}^{\ast }$ are divisors of $31$$31$3131$31$, which are $1$$1$11$1$ and $31$$31$3131$31$ itself, because $31$$31$3131$31$ is a prime number.

Therefore, while an element $\alpha$$\alpha$alpha\alpha$\alpha$ with order $31$$31$3131$31$ (a primitive element) will indeed generate the entire field ${\mathbb{F}}_{2^5}$${\mathbb{F}}_{2^5}$F_(2^(5))\mathbb{F}_{2^5}${\mathbb{F}}_{2^5}$ through its powers, implying ${\mathbb{F}}_{2^5}={\mathbb{F}}_2[\alpha ]$${\mathbb{F}}_{2^5}={\mathbb{F}}_2[\alpha ]$F_(2^(5))=F_(2)[alpha]\mathbb{F}_{2^5} = \mathbb{F}_2[\alpha]${\mathbb{F}}_{2^5}={\mathbb{F}}_2[\alpha ]$ for such $\alpha$$\alpha$alpha\alpha$\alpha$, not every non-identity element has this property, as the statement incorrectly generalizes. Only the primitive elements of ${\mathbb{F}}_{2^5}$${\mathbb{F}}_{2^5}$F_(2^(5))\mathbb{F}_{2^5}${\mathbb{F}}_{2^5}$ can generate the entire field.

The statement is false because not every non-identity element in ${\mathbb{F}}_{2^5}$${\mathbb{F}}_{2^5}$F_(2^(5))\mathbb{F}_{2^5}${\mathbb{F}}_{2^5}$ is a primitive element capable of generating the entire field. Only specific elements, namely the primitive elements, have this property.