I'm trying to solve the following problem:  Let $p\geq 3$ be a prime number, let $r \in \Bbb N$, and let $x$ be a primitive root modulo $p^r$. Show that $x$ is a primitive root modulo $p$.    I'm pretty much out of ideas. I tried to use the following claim: $a$ is a primitive root modulo $n$ if and only if for every  prime $q$ such that $q$ divides $\varphi(n)$ we have: $$a^{\frac{\varphi(n)}{q}}\not\equiv 1\pmod n$$  But got nothing.
For an odd prime $p$, prove that any primitive root $r$ of $p^n$ is also a primitive root of  $p$   So I have assumed $r$ have order $k$ modulo $p$ , So $k|p-1$.Then if I am able to show that $p-1|k$ then I am done .But I haven't been able to show that.Can anybody help me this method?Any other type of prove is also welcomed.
What is the logarithm of a kilometer? Is it a dimensionless number? In log-plots a quantity is plotted on a logarithmic scale. This got me thinking about what the logarithm of a unit actually is.  Suppose I have something with length $L = 1 \:\mathrm{km}$.  $\log L = \log \mathrm{km}$  It seems that the unit of $\log L$ is $\log \mathrm{km}$, but I can also say $L = 1000 \mathrm{\:m}$ and now:  $\log L = 3 + \log \mathrm{m}$  This doesn't appear to have any units at all.  This suggests that $\log \mathrm{km}$ and $\log \mathrm{m}$ are actually dimensionless numbers. But wait, I can do this with any unit! Does it actually make sense to talk about the logarithm of a unit, or some other function for that matter? Two Answers, One Question Sometimes, when I answer a question on Stack Overflow, I have two different answers that are equally valid. This happens most when answering architectural questions where there are many valid ways to solve a problem.  Is it better to put each alternative into a single answer, or put each alternative in its own answer?  Posted each alternative separately will allow the voting system to float the best answer to the top, but I don't want to be accused of rep whoring by "flooding" a question with answers.  What's your opinion? One or two answers? Sufficient condition(s) for $f'$ to be Riemann integrable Please rate and comment. I want to improve; constructive criticism is highly appreciated. Please take style into account as well.  The following proof is solely based on vector space related axioms. Axiom names are italicised. They are defined in Wikipedia (see vector space article).  Vector spaces - Multiplying by zero scalar yields zero vector  \begin{array}{lrll} \text{Let} &amp; \dots &amp; \text{be} &amp; \dots \\ \hline &amp; F &amp;&amp; \text{a field.} \\ &amp; V &amp;&amp; \text{a vector space over $F$.} \\ &amp; 0 &amp;&amp; \text{an identity element of addition of $F$.} \\ &amp; \mathbf{0} &amp;&amp; \text{an identity element of addition of $V$.} \\ &amp; \mathbf{v} &amp;&amp; \text{an arbitrary vector in $V$.} \\ \end{array}  $$\text{Then, }0\mathbf{v} = \mathbf{0}.$$  Proof. We will denote by $1$ an identity element of scalar multiplication; we will denote by $(-\mathbf{v})$ an additive inverse of $\mathbf{v}$. \begin{align*} 0\mathbf{v} &amp;= 0\mathbf{v} + \mathbf{0}                     &amp;&amp; \text{by }\textit{Identity element of vector addition} \\ &amp;= 0\mathbf{v} + (\mathbf{v} + (-\mathbf{v}))   &amp;&amp; \text{by }\textit{Inverse elements of vector addition} \\ &amp;= (0\mathbf{v} + \mathbf{v}) + (-\mathbf{v})   &amp;&amp; \text{by }\textit{Associativity of vector addition} \\ &amp;= (0\mathbf{v} + 1\mathbf{v}) + (-\mathbf{v})  &amp;&amp; \text{by }\textit{Identity element of scalar multiplication} \\ &amp;= ((0 + 1)\mathbf{v}) + (-\mathbf{v})          &amp;&amp; \text{by }\textit{Distributivity of scalar multiplication (field addition)} \\ &amp;= ((1 + 0)\mathbf{v}) + (-\mathbf{v})          &amp;&amp; \text{by }\textit{Commutativity of field addition} \\ &amp;= (1\mathbf{v}) + (-\mathbf{v})                &amp;&amp; \text{by }\textit{Identity element of field addition} \\ &amp;= \mathbf{v} + (-\mathbf{v})                   &amp;&amp; \text{by }\textit{Identity element of scalar multiplication} \\ &amp;= \mathbf{0}                                   &amp;&amp; \text{by }\textit{Inverse elements of vector addition} \\ \end{align*} QED Coordinating conjunction immediately followed by parenthetical — Is a comma needed or not? states:     If a dependent clause, or an introductory phrase requiring to be set off by a comma, precedes the second independent clause, no comma is needed after the conjunction.   Strunk's example illustrates the point:     The situation is perilous, but if we are prepared to act promptly, there is still one chance of escape.   However, I would be tempted to put a comma after the "but," like this:     The situation is perilous, but, if we are prepared to act promptly, there is still one chance of escape.   So, I know you're thinking, "Well, Strunk's Rules of Usage says that you're wrong. No comma."   But is Rules of Usage really the best source? I'd like to see this rule in a more contemporary and authoritative source, like the Chicago Manual of Style.  There are many instances —  — where the comma is included before a parenthetical comment that comes after a coordinating conjunction.  Any authoritative source on the issue? Can I force an iPhone user to upgrade an application? Separate selected UV map vertices from surrounding vertices How do I break the selected UV map from its surrounding vertices? Driver locked memory on a non-virtual machine Prove that $\lim_{x \to \infty}\big(\frac{x}{x-1}\big)^x$ is also $e$. Trying to make sense out of the idea that $100\%$ continuous decay is $\frac{1}{e}$, I thought about this:  You can express $1+\frac{1}{x}$ as $\frac{x+1}{x}$, such that $\big(1+\frac{1}{x}\big)^x = \big(\frac{x+1}{x}\big)^x$  And you can express $1-\frac{1}{x}$ as $\frac{x-1}{x}$, such that $\big(1-\frac{1}{x}\big)^x = \big(\frac{x-1}{x}\big)^x =\frac{1}{\big(\frac{x}{x-1}\big)^x}$  Now $\big(\frac{x}{x-1}\big)^x$ and $\big(\frac{x+1}{x}\big)^x$ look very similar, and I can imagine (and see on Mathematica) that $\lim_{x \to \infty} \big(\frac{x}{x-1}\big)^x$ also approaches $e$.  However I'm clueless about limit proofs, how do you prove that?  EDIT: Just had a last second insight.  Can you say that $$\lim_{x \to \infty} \big(\frac{x}{x-1}\big)^x = \lim_{x \to \infty} \big(\frac{x}{x-1}\big)^{x-1} \cdot \lim_{x \to \infty} \big(\frac{x}{x-1}\big)$$ $$= \lim_{x \to \infty} \big(\frac{x}{x-1}\big)^{x-1} \cdot 1$$ $$=\lim_{x \to \infty} \big(\frac{x+1}{x}\big)^x = e ?$$   (not sure about that last limit transition...I mean I'm pretty sure it's true, just not sure how to write it; I would appreciate feedback on that, thank you) free groups: $F_X\cong F_Y\Rightarrow|X|=|Y|$ I'm reading Grillet's Abstract Algebra. Let $F_X$ denote the free group on the set $X$. I noticed on wiki the claim $$F_X\cong\!\!F_Y\Leftrightarrow|X|=|Y|.$$ How can I prove the right implication (find a bijection $f:X\rightarrow Y$), i.e. that the rank is an invariant of free groups?  I am hoping for a simple and short proof, having all the tools of Grillet at hand. Rotman (Advanced Modern Algebra, p.305) proves it only for $|X|&lt;\infty$, Bogopolski's (Introduction to Group Theory, p.55) proof seems (unnecessarily?) complicated, and Lyndon &amp; Schupp's (Combinatorial Group Theory, p.1) proof I don't yet understand. It's the very first proposition in the book; in the proof, they say:     The subgroup $N$ of $F$ generated by   all squares of elements in $F$ is   normal, and $F/N$ is an elementary   abelian $2$-group of rank $|X|$. (If   $X$ is finite, $|F/N|=2^{|X|}$ finite;   if $|X|$ is infinite, $|F/N|=|X|$). $\square$   Is $N:=\langle w^2;w\in F\rangle$? What is an abelian $2$-group? Elementary? What and how does the above quote really prove? I'm guessing a free abelian group on $X$ is $\langle X|[X,X]\rangle\cong\bigoplus\limits_{x\in X} \mathbb{Z}$?  Can an isomorphism $\varphi:F_X\rightarrow F_Y$ not preserve the length of words? At least one letter words? Is every Lie subgroup of $SU(2)$ closed? Why are two different concepts both called "heap"? How to discard unsaved changes to a view when editing it leads to a WSOD? Is there a word for describing people who know many things but superficially? Is there an appropriate word for the antonym of "full-stack", as in programming? Someone who knows a lot of things just as a full-stack programmer, but without mastering them. Why doesn't planet Earth expand if I accelerate upwards when standing on its surface? According to General Relativity I am being accelerated upwards by planet earth while writing this question. But a curious person on the the other side of the planet relative to me would have the same experience. That means we are accelerated in opposite directions, although earths diameter do not seem to increase. How can this be? There is an exercise in Humphreys's An Introduction to Lie Algebras and Representation Theory: "Any nilpotent Lie algebra contains a codimension 1 ideal".  The proof I am thinking of is the following.  Suppose the Lie algebra $L$ satisfies $L\neq[L,L]$. Then $\pi\colon L\to L/[L,L]$ is a projection onto an abelian algebra. Choose a codimension 1 subspace $I\subset L/[L,L]$. Then $I$ is an ideal and its preimage $\pi^{-1}(I)$ is necessarily an ideal in $L$, because any subspace containing the derived algebra is an ideal. Moreover, $\pi^{-1}(I)$ has codimension 1. $\square$  The only condition on $L$ is that $L\neq[L,L]$, which holds for solvable Lie algebras as well. Is this proof correct? How come Humphreys didn't set the exercise with "nilpotent" replaced by "solvable"? Floating sidebar clash with footer in team page Here is a screenshot from  page, with the floating sidebar: (all good so far)        After I scroll down to footer it just combines both:        Can it please be fixed? Why doesn't Ubuntu 16.04 in Windows have ubuntu.exe? What is the value of the series   $\sum\limits_{n=1}^\infty {\frac{1}{10^n-1}}$   and can one find closed forms for the sums   $\sum\limits_{n=1}^N {\frac{1}{10^n-1}}$?   I found that $\sum_{n=1}^N {\frac{1}{10^n-1}} $  =  $\sum_{n=1}^N (\lim_{m\to\infty  }{\frac{10^{-n}-10^{-nm}}{1-10^{-n}}})$  =  $0.111\ldots+0.0101\ldots+\ldots$ but I can't go farther.     The original question is: "Where does the second '1' appear?" A Light is shone out the side window of a spaceship ( moving close to light speed ), will the light still travel at the speed of light in the B axis?