I can see why there is only one element in $GF(p) \setminus \{0\}$ of order 2. But I'm not sure why I can claim that this would be the same element and only element for $GF(p^r)$.
122674
In  field of characteristic $p$, where $p$ is prime, why can $x^2=1$ only have the two solutions $1$ and $-1$? Is it just because that polynomial only has degree 2?
57918
Suppose $n = p_1^{a_1} ... p_r^{a_r}$ where $p_i$'s are prime integers and $a_i$'s are positive integer.  (a) Show that the set of nilpotent elements in $\mathbb{Z}_n$ is $(p_1 p_2 ... p_{r-1} p_r)$, the ideal generated by the element $p_1p_2 ... p_{r-1} p_r$.  (b) Find a necessary and sufficient condition on $n$ such that $\mathbb{Z}_n$ does not have any non-zero nilpotent elements.  My attempt:  (a)Let x be a nilpotent element in $\mathbb{Z}_n$. Then $x^k = n$ for some $k &gt; 0$.Note that $x \neq 1$. Since $Z_n$ is a UFD, $x =  p_1^{b_1} ... p_r^{b_r}$, where $b_i &gt; 0$ and so $x^k = p_1^{k b_1} ... p_r^{k b_r} = p_1^{k b_1 - 1} ... p_r^{k b_r - 1}(p_1 ... p_r) \in (p_1 ... p_r)$. Hence the set of nilpotent elements is a subset of $(p_1 ... p_r)$. On the other hand, for all $x \in (p_1 ... p_r)$, $x = z_1 \cdot p_1 ... p_r \cdot z_2$ for some $z_1 = p_1^{c_1} ... p_r^{c_r}, z_2 = p_1^{d_1} ... p_r^{d_r} \in \mathbb{Z}_n$, so $x = p_1^{c_1 + d_1 +1} ... p_r^{c_r + d_r+1}$. Let k be an natural number such that $k(c_i+d_i+1) &gt;a_i$ for all i. Then we have $x^k = p_1^{k(c_1 + d_1 +1)} ... p_r^{k(c_r + d_r+1)} = p_1^{a_1} ... p_r^{a_r}(p_1^{k(c_1 + d_1 +1)-a_1} ... p_r^{k(c_r + d_r+1)-a_r}) = 0 \in\mathbb{Z}_n$. Hence the $(p_1 ...p_r) \subseteq$ the set of nilpotent elements in $\mathbb{Z}_n$. Therefore $(p_1 ...p_r) =$ the set of nilpotent elements in $\mathbb{Z}_n$  (b) $a_i = 1$ for all $i$. If $a_i = 1$ for all i, then it is obvious then $\mathbb{Z}_n$ does not have any non-zero nilpotent elements. Now I just need to show that if $\mathbb{Z}_n$ does not have any non-zero nilpotent elements, then $a_i = 1$ for all $i$. Proof by contradiction: Suppose there exists some $a_i$ > 1. Then $p_1...p_r &lt; n$ is non-zero a nilpotent element.
48711
Let $G$ be a non-trivial cyclic group with an unique generator. Show that $G$ has exactly two elements.   Let $g$ be a generator of $G$. My first idea to approach the problem was to show that if $|G|=n &gt; 2$ then there is an integer $k$ between 1 and $n$ such that $\langle g^k \rangle = G$ or, equivalently, $g \in \langle g^k \rangle$. My second idea was to analyse the problem in 2 separate cases. For the first one, if $n$ is prime, let $k$ be a positive integer then, by Lagrange's Theorem, $|g^k| \mid n$ and, therefore, $\langle g^k \rangle = n$. Hence, we must conclude that, if $n &gt; 2$ the generator cannot be unique. For the second case, where $n$ is composite, I was hoping to get to a contradiction, but could not advance any further. Is there a way to develop either of these ideas or is it necessary a new approach?
31255
If $d$ is a positive divisor of $n$, the number of elements of order $d$ in a cyclic subgroup of order $n$ is $\phi (d)$=the number of positive natural numbers less than $d$ which are coprime to $d$.  The question I have concerns a part of the proof:  If $d | n$ then there exists exactly one subgroup of order $d$ -- call it $\langle a \rangle$.  Then every element of order $d$ also generates the subgroup $\langle a \rangle$ and an element $a^k$ generates $\langle a \rangle$ iff $gcd(k,d) = 1$ implies that the number of such elements is precisely $\phi (d)$.  How does every element of order $d$ also generate the subgroup $\langle a \rangle$, wouldn't it be only one $a$ since $|\langle a \rangle| = |a|$?  And how does this fact imply $\phi(d)$ is the correct number?
31860
I would like to construct a formal proof of the following:  $$(A\setminus C)\setminus(B\setminus C) = (A\setminus B)\setminus C$$  Let $a∈A$ be an arbitrary element, we will show that $a\in A \cap \overline B \cap \overline C$.  For LHS, since $a\in A$, we have that $a\in (A \cap  \overline C) \cap  (\overline B \cap \overline C)$. This is equivalent to $a\in (A \cap  \overline B) \cap \overline C$.    For RHS, since $a\in A$, we have that $a\in (A \cap  \overline B) \cap \overline C$  $\therefore lhs \equiv rhs$ and this concludes the proof  I would be grateful for any feed back on this element chasing proof. Is it flawed or where should improvements be made?  Thanks
56397
Prove that the order of an element in $S_n$ equals the least common multiple of the lengths of the cycles in its cycle decomposition.  Proof: Let $\sigma \in S_n$. Then $\sigma = (a_1a_2...a_{m_1})(a_{m_1+1}a_{m_1+2}...a_{m_2})....(a_{m_{k-1}+1}a_{m_{k-1}+2}...a_{m_{k}})$ represents the cycle decomposition of $\sigma$. Suppose $|\sigma| = n$ is the order of an element in $S_n$. So $\sigma^n = 1$. Can someone please help  me?  I don't know how if I am doing this fine. And I am stuck. Thank you for any help.
106485
Let $G$ be a finite group and suppose there exists $f\in\text{Aut}(G)$ such that $f^2=\text{id}_G$, i.e., $f$ is its own inverse, and such that $f$ has no fixed points other than the identity $e$ of $G$, i.e., $f(x)=x\Rightarrow x=e$. Show that $G$ is necessarily abelian.   While trying to do this exercise I noticed two facts.  First, $g$ and $f(g)$ have the same order because $o(f(g))|o(g)$ and, applying $f$ again and using $f^2=\text{id}_G$, $o(f(f(g)))=o(g)|o(f(g))$ and, once the order of any element is $\geq 1$, it follows that $o(g)=o(f(g))$. Also, it's easy to see that $g$ and $f(g)$ commute.  Second, there cannot exist such an automorphism if the order of $G$ is even, because $f(e)=e$ and we can form pairs like $\{g,f(g)\}$ with $f(g)\neq g$ that are invariant under $f$, i.e., $f(\{g,f(g)\})=\{g,f(g)\}$. But once the order of $G$ is even, proceeding with the construction of the pairs, we'll end up with just one element $\neq e$, so we must have some $\gamma\in G\setminus{e}$ with $f(\gamma)=\gamma$, contradicting the hypothesis.  Let $n$ be the order of $G$ and let's fix an enumeration of the elements of $G$, say $G=\{g_1,\ldots,g_n\}$. My approach was the following. For each $\sigma\in S_n$, let $x_{\sigma}=\prod_{i=1}^ng_{\sigma(i)}$. Then we have that $f(x_{\sigma})=x_{\sigma^{\prime}}$. If it's shown that $f(x_{\sigma})=e\;\forall\;\sigma\in S_n$, then we'd have $x_{\sigma}=e\;\forall\;\sigma\in S_n$, which implies that $G$ is abelian. The problem is that $S_n$ has $n!$ elements while $G$ has $n$ elements, so there repetitions among the $x_{\sigma}$, and we cannot apply directly the reasoning of the last paragraph.  Is this the right way, or is there an easier manner to solve this?
12598
I was trying to prove this question.   Q. If G is a group of even order, prove that it has an element a, such that a$\neq$e satisfying $a^2$=e.   I tried solving it using Principle of mathematical induction.  So, Let O(G)=2m, where m$\in$ $N$.   We proceed by induction on m.   For m=1, O(G)=2, $\therefore$ G contains two elements.  G=$\{{e,a}\}$. Now, e has order 1. So, a can not have order 1. $\therefore$ a has order 2. So, it true for m=1.  Now, let it be true for m=k.  Then, we have a group G of order 2k, where $\exists$ an element a$\in$ G, whose order is 2.   We wish to prove it for m=k+1.   Then G has order 2k+2.   I don't know how to proceed further.
17078
Question :   If $|G|=pq$ where $p$ and $q$ are primes that are not necessarily distinct.   Prove that the order of $Z (G) =1$ or $pq$.   Showing the order is $pq$ is trivial.   I unsure how to start with showing the order is $1$. Hints are appreciated.  Thanks in advance.
121335
GF(q) is a finite field of order q, where q is odd. Prove that $a\in GF(q), a\neq0$ has a root in $GF(q)$ iff $a^{(q-1)/2}=1$.  I tried to prove it this way:  Suppose a has a root in $GF(q)$, so there is some $b\in GF(q)$ so that $b^2=a$. There is a primitive element $g$ in $GF(q)$, so that $g^x=a, g^y=b$. Therefore $(g^y)^2=g^x\to 2y=x$. Furthermore $a^{(q-1)/2}=(g^x)^{(q-1)/2}=g^{(x/2)*(q-1)}=g^{y(q-1)}$ and since $y$ is an integer, $g^{y(q-1)}=1^y=1$. Now suppose $a^{(q-1)/2}=1$, then again $(g^x)^{(q-1)/2}=1\to g^{(x/2)*(q-1)}=1$. That means that $x/2$ is an integer (or else $g^{(x/2)*(q-1)}\neq 1$), and therefore $0\le x/2\le q-1$ being an integer, $g^{x/2}$ creates a certain element $b\in GF(q)$ so that $b^2=a$.  I am a bit unsure of this proof - first of all since I don't use the fact that $q$ is odd, which means I am wrong somewhere. Second of all, I feel extremely 'weak' with my argument regarding $x/2$ being an integer, yet I have no idea how to strengthen it. Any hints/directions will be extremely appreicated!
71320
How do i show that $D_6$ has an element of order 6 but $A_4$ has only elements of order 3 and 2 and thus,  $A_4$ is not isomorphic to $D_6$? Can someone explain to me? Thanks!
107055
Prove or disprove:  The set: $\{ \emptyset\}^{\Bbb N}$ has only one element.  The set $\emptyset^{\Bbb N}$ is empty.    I'm pretty sure both are true, for 1. all the natural numbers go to the empty set and that is the only element. Where as 2. all the natural numbers go to nothing so that's why it's empty. (Same as $ {\Bbb N}^{\emptyset} $ nothing goes to something is nothing (I think)). The problem is I have no idea how to show this so advice would be appreciated.  Edit: My defintion of a function is: $f:B\to A$ if  $(a_1,b),(a_2,b)\in f \ \Rightarrow \ a_1=a_2$ So for 2. I can simply say that: $(1,\emptyset),(2,\emptyset)\in f \Rightarrow a_1\neq a_2$  ?
79950
$G$ is a Group of order $pq$, if $G$ has exactly one subgroup of order $p$ and another with order $q$, then $G$ is cyclic ($p,q$ are prime)  Here's what I have so far, I'm not sure if it is correct.  Let $H, K$ be subgroups of $G$ $|H| = p$  $|K| = q$   We know that by Lagrange's Theorem:  $\forall a \in H : |a|=p$ where $a \neq e$ $\Rightarrow H = &lt;a&gt;$  Similarly: $\forall a \in K : K = &lt;a&gt;$  We also know that: $\forall a \in G : |a| | pq \Rightarrow |a| = p$ or $|a| = q$   $|a| = p \Rightarrow a \in H$ since there is exactly one subgroup of order $p$  $|a| = q \Rightarrow a \in K$ for similar reasons  This means every element in $G$ is either in $H$ or $K$, which are both cyclic. That means $G$ is cyclic.  What am I missing here? Have I made any mistakes and/or is the proof incomplete?
122412
Suppose that $G$ is a group that has exactly one nontrivial proper subgroup. Then we have to show that $G$ is cyclic and order of $G$ is $p^2$ where $p$ is prime.   I tried as, if $a$ and $b$ two element of $G$ of order $p$ and $q$ then $\langle a \rangle$ and $\langle b \rangle$ are two proper subgroups – a contradiction. Thus every element of $G$ have order some power of $p$. After this I am stuck. Thanks for help.
61307
Since $T_p G$ is isomorphic to $T_e G$ for all $p\in G$, it makes sense that each vector in $T_p G$ can be identified with a vector in $T_e G$. Hence, to make the map from $TG$ one to one, we must identify each vector in $T_e G$ with a point in $G$, but I'm having trouble saying this precisely.  Thanks so much!
117486
This question is a slight generalization of .  How many elements are there in a finite field of order $q$ which are :    Squares. Cubes. Higher powers.   I mean :  How many elements are there which are squares...   How many elements are there which are cubes...   how many elements are there which are higher powers...   What all I could see is  :  For finding elements which are cubes I would consider :  $\eta : F^*\rightarrow F^*$ sending $x\rightarrow x^3$  For this If I know the kernel of $\eta$ then I would have $F^*/Ker(\eta) \cong \{x^3 : x\in F^*\}$  Which says that :  No of elements which are cubes are $\dfrac{|q-1|}{|Ker (\eta)|}$.  Now, Kernel of $\eta$ would be $\{x\in F^* : x^3=1\}$  All I know is if there is some subgroup (I am mentioning $\{1,x,x^2\}$ for $\{x\in F^* : x^3=1\}$)  Then It would be unique as a finite cyclic group can not have two subgroups of same order.  So, Now the problem is how to see for existence as we are through with uniqueness.  Suppose I prove uniqueness then I would say :  Number of elements of $F^*$ which are squares are :   $\dfrac{q-1}{3}$ If $F^*$ have an element of order $3$  $q-1$ If $F^*$ have no  element of order $3$    (I believe this would be totally dependent of nature of $q$)  Now the question is how do i make sure of existence with given nature of $q$.  I would be Thankful If some one can help me to see this and I would be happy to see further generalization.     How many elements of order $n$ are there in a Finite field of cardinality $q$   Thank you :)
49324
Suppose $A$ is the set of all rational numbers $p$ such that $p^2 &lt;2$ and $B$ is the set of all rational numbers $p$ such that $p^2 &gt; 2$. We want to show that $A$ contains no largest element and $B$ contains no smallest element.  In Rudin's Principles of Mathematical Analysis, he associates $q = p- \frac{p^2-2}{p+2} = \frac{2p+2}{p+2}$. Did he just come up with this using trial and error?