Assume $G = \{1, a,b,c\}$ is a group of order $4$ with identity $1.$ Assume also that $G$ has no elements of order $4$. Show that there is a unique group table for $G$. Also show that $G$ is abelian.   If $G$ is abelian, then the group table matrix must be symmetric. How can I introduce a binary function and show it? I am new in this field, so I am not so familiar. I have proved many other exercises, but it is a little tough (for me).   Can you please help?  Edit: I know every element has order $\leq 3$ , but I do not understand how I will proceed.
Not sure how to make tables, but:  For a binary operation $*$, and set $\{e,a,b,c\}$, in the Cayley table, $a*a$ can be filled with either the identity or an element different from both $e$ and $a$. If in the table the place for $a*a$ is filled with $e$, then the rest of the table can be filled out in two different ways.  What are the two different ways?  Thanks
No homomorphism from $Z_{16}\oplus Z_{2}$ onto $Z_{4}\oplus Z_{4}$. For some reasons, homomorphism is a very hard area for me to make improvements. I've been hitting the brick wall for almost 2 hours.   Prove that no homomorphism exists from $Z_{16}\oplus Z_2$ onto $Z_4\oplus Z_4$.  Assume a homomorphism $\phi$ exists from $Z_{16}\oplus Z_2$ onto $Z_4\oplus Z_4$. Then $\phi$ is an isomorphism.  Note that the $\left | \ker \phi \right |=2$ The $\ker \phi$ is also a normal subgroup for $Z_{16}\oplus Z_{2}$. We want possible normal subgroup of order 2. I.e., 2 elements in each normal subgroups.  By Lagrange's theorem, the order of each element in a group divides the order of a group so the possible order of the elements are 1 or 2. If the elements has order 1, then the Ker \phi cannot have 2 elements.  Hence, $\ker\phi$ = $\left \{ (0,0),(8,0) \right \}$ ,$\left \{ (0,0),(0,8) \right \}$,$\left \{ (0,0),(8,1) \right \}$, possibly.  By the First isomorphism theorem:  $\phi: Z_{16}\oplus Z_2 \rightarrow Z_2\oplus Z_2$  $\Psi: G/\ker \phi \rightarrow \phi\left ( Z_16\oplus Z_2 \right )=Z_2\oplus Z_2$  $g\ker \phi  \mapsto \phi(g)=\Psi(g\ker \phi)$  Any help is appreciated.  Edit: I know that $(2,0)$ has order $8$ in $Z_{16}\oplus Z_{2}$ but any elements in $Z_{4}\oplus Z_{4}$ does not have order $8$ which would have solved the question at the outset. But I would like a different route. Why are there no Pokémon near me? Define additional math operators to be typeset in roman Mathematical operators, such as function names, should be set in roman type, not italics. Latex already has commands for some operators, including \max, \min, and \log.  How can I define additional such commands? Gollum had his life largely extended because he possessed the One Ring. He appeared to be an old, deformed, and twisted creature.  According to , he had the Ring for 478 years. His loss of the Ring and the events of The Lord of the Rings are separated by around 77 years.  Did the Ring continue to extend his life after he lost it? I am trying to prove the following statement but have trouble  comprehending/going forward with some parts! Here is the statement:     If $A$ is any set, then  $|A|$ $&lt;$ $|P(A)|$   Here is what I have so far:  We need to show that there is an injection from $A$ to $P(A)$ but not a surjection.   A natural choice for an injection is the function $ f(x)$ $=$ $\{x \}$, which in plain English, takes any element $x$ (that is in $A$) and sends it to the one-element set $\{x \}$. Thus $f(x)$ is injective!  To show that there is no surjection, for the sake of contradiction, assume there is a surjection. Here is where I start to have trouble. Surjectivity means that every element of the co-domain is mapped to an element of the domain, correct? Consequently, in this particular case, we are "matching" sets (from $P(A)$) to elements (from $A$) right?  If the above is correct, my problem arises here. I am not sure how to prove that $f$ is not surjective. Unfortunately, I am easily confused by notation so please explain in English.  Thank you in advance!! :) I've been struggling with this all day today. I imagine it's not very hard, but my algebra skills are terrible. So, how can I show that if $m&gt;n$ and $a$ is a positive integer, then $$a^{2^n}+1 \mid a^{2^m}-1.$$ I just can't get a coherent picture of what I'm supposed to do. If this too (see yesterday's question) boils down to the Unique Factorization Theorem, I would be very grateful for any intuition on how to think about this! What is taking up so much space on my disk, beside the filesystem? I have only one disk in my computer, a 80 GB SSD. It is formatted as a single ext4 partition (no swap), and all the usual folders are installed on it (I keep lots of data on external media, but /home and all the rest is on the SSD).   Today I booted it and I got a message that the drive is full. I opened Disk Usage Analyzer to look at what takes up space. It insists that 67.8 GB of the disk are used, and that / is taking up 36.4 GB. Which leads to the question, where are the missing 30 GB, if not on the file system?!    Just to have a comparison, I added up all the sizes of all readable folders listed in Nautilus as subfolders of / (including the hidden ones). I got 20.9 GB. Trash was unreadable, but I know it has 16.2 GB, so the sum is 36.1 GB, around the same as what Disk Usage Analyzer is reporting. There were some system folders which were unreadable, like proc, but I doubt that they add up to 30 GB - else it wouldn't be possible to install Ubuntu on small disks, and I've seen it run on 2.5 GB. I think they must be the 0.3 GB difference between my calculation and the Disk Usage Analyzer's report.   So I'd like to know, what is eating up these 30 GB and how do I get it to free them?     Edit with answers to CYREX's questions   Booted from Ubuntu 10.10 64-bit live CD (same as my system).  fsck result:   root@ubuntu:~# fsck /dev/sda1 fsck from util-linux-ng 2.17.2 e2fsck 1.41.12 (17-May-2010) /dev/sda1: clean, 265956/4890624 files, 18073343/19537408 blocks  Booted from the SSD. Emptied the trash, and the 16 GB from there are now free. The 30 GB are still missing.    I'm not experienced enough to know what counts as weird in a log. Here are my messages log and my syslog since last boot. Maybe you can find something in them:       Here it got really strange. I plugged in an external 500 GB HDD. Disk Usage Analyzer overreported the available space, then showed space missing (76 GB used, but only 24 in folders).      Booted again from the LiveCD, started the Disk Usage Analyzer from it and got the same results as from the installed Ubuntu, within 1-2 GB.   Edit with answer to CodeMonk:  That would have been a nice solution, but the partition is really 80 GB - I mean 74 GB + marketing "error", anyway the whole disk is formatted. It also shows that over 50 GB are used, so where are the 30 GB if not in files and folders?    GParted also reports the correct size of the external HDD. Is There a Natural Way to Extend Repeated Exponentiation Beyond Integers? This question has been in my mind since high school.  We can get multiplication of natural numbers by repeated addition; equivalently, if we define $f$ recursively by $f(1)=m$ and $f(n+1)=f(n)+m$, then $f(n) = m \times n$.  Likewise, we get exponentiation by repeated multiplication. If $g(1)=m$ and $g(n+1)=mg(n)$, then $g(n) = m^n$.  In my high school mind it was natural to imagine a new function defined by repeated exponentiation: $h(1)=m$ and $h(n+1)=m^{h(n)}$.  These definitions only make sense for $n$ a natural number, but of course there are standard very mathematically satisfying ways to define multiplication and exponentiation by any real number. My question is this:  Can the function $h$ defined above also be extended in a natural way to $\mathbb{R}^{&gt;0}$?  The question is in the spirit of seeking an extension of $f(n)=n!$ to $\mathbb{R}$ and arriving at $\Gamma(x)$.  Let me focus the question, and attempt to make precise what I mean by "in a natural way." Take $h(1)=2$ and $h(n+1)=2^{h(n)}$. $h$ is now defined on $\mathbb{N}$, and $h(2)=4$, $h(3)=16$, $h(4)=2^{16}=65,536$ etc. Is it possible to extend the domain of definition of $h$ to all positive reals in such a way that  a) The functional equation $h(x+1)=2^{h(x)}$ continues to be satisfied for all $x$ in the domain.  b) $h$ is $C^\infty$. (Analytic would be even better but this seems maybe too much to hope for?)  c) All $h$'s derivatives are monotone.  These requirements are my attempt to codify what would count as "natural." I am open to suggestions about what would be a better list of requirements.  If such a function exists, I would like to know how to construct it; if it doesn't, I would like to know why (i.e. outline of proof), and if relaxing some of the requirements (e.g. just the first derivative monotone) would make it possible.  (If the function exists, I am also interested in the questions, "is it unique?" "Could we add some natural requirements to make it unique?" But my main query is about existence.) The following exercise comes from Fulton's Algebraic Curves:     Let $k$ be an infinite field and $V = \{(t,t^2,t^3) \, ; \, t \in k \} \subset \mathbb{A}^3(k)$. Show that $V$ is algebraic and compute $I(V)$ by giving a finite set of generators.   Now, showing $V$ is algebraic is easy by noting that $V = V(X^2-Y, X^3-Z)$, but I can't find a way to describe $I(V)$ explicitly. Obviously $(X^2-Y)(X^3-Z) \in V$, but how can I find the actual generators of this ideal? Let $G$, $H$, and $K$ be abelian groups. Assume $G \oplus H \cong G \oplus K$. If each are finitely generated, then the structure theorem easily gives us that $H \cong K$. However, there are counterexamples in the general case, namely $H=\mathbb{Z}$, $K=\mathbb{Z} \oplus \mathbb{Z}$, and $G=\bigoplus_{i=1}^\infty\mathbb{Z}$.   I am wondering what goes wrong with the following argument:   Form the split short exact sequence  $$0 \to G \xrightarrow{i} G \oplus H \xrightarrow{\pi} H \to 0$$  where $i$ and $\pi$ are the inclusion and projection, respectively. Then don't we immediately get from the first isomorphism theorem that $H \cong (G \oplus H)/G$, which would contradict the counterexample above?   If we run this for the counterexample given, we get the sequence   $$0 \to \bigoplus_{i=1}^\infty\mathbb{Z} \to \bigoplus_{i=1}^\infty\mathbb{Z} \to \mathbb{Z} \to 0$$  which seems to still be short exact, even though here $G \cong G \oplus H$. But of course here we can see directly that $(G \oplus H)/G \cong G/G$, which is trivial. So what went wrong?   It seems to me to be something along these lines: In this case, the kernel of the projection is not the same as the original group, it is just, say, $0 \oplus \mathbb{Z} \oplus \mathbb{Z} \oplus \dots$, i.e. nontrivial from the second coordinate on, which just happens to be isomorphic to the original group. So if we form the other short exact sequence with $K$ in place of $H$, then the difference seems to be that we have two different projection maps with different (but isomorphic) kernels.   But I'm not really sure how to explain this in the general case, if my reasoning is correct to begin with.   Thanks in advance for any help! Best way to find a point on a circle closest to a given point Given a point (pX, pY) and a circle with a known center (cX,cY) and radius (r), what is the shortest amount of code you can come up with to find the point on the circle closest to (pX, pY) ?  I've got some code kind of working but it involves converting the circle to an equation of the form (x - cX)^2 + (y - cY)^2 = r^2 (where r is radius) and using the equation of the line from point (pX, pY) to (cX, cY) to create a quadratic equation to be solved.  Once I iron out the bugs it'll do, but it seems such an inelegant solution. Is it possible to change the linkage of Nether portals? How to customize the Ubuntu Live CD? Intuition behind logarithm change of base I try to understand the actual intuition behind the logarithm properties and came across  that explains the multiplication and thereby also the division properties very nicely:     Suppose you have a table of powers of 2, which looks like this: (after revision)      $$\begin{array}{rrrrrrrrrr} 0&amp;1&amp;2&amp;3&amp;4&amp;5&amp;6&amp;7&amp;8&amp;9&amp;10\\ 1&amp;2&amp;4&amp;8&amp;16&amp;32&amp;64&amp;128&amp;256&amp;512&amp;1024 \end{array}$$      Each column says how many twos you have to multiply to get the number in that column.  For example, if you multiply 5 twos, you get $2\cdot2\cdot2\cdot2\cdot2=32$, which is the number in column 5.      Now suppose you want to multiply two numbers from the bottom row, say $16\cdot 64$.  Well, the $16$ is the product of 4 twos, and the $64$ is the product of 6 twos, so when you multiply them together you get a product of 10 twos, which is $1024$.   I found that very helpful to understand the actual proofs for this property.  I still struggle to get the idea behind the change of base rule. I'm familiar with the proof that goes like:  $$\log_a x = y \implies a^y = x$$ $$\log_b a^y = \log_b x$$ $$y \cdot \log_b a = \log_b x$$ $$y = \frac{\log_b x}{\log_b a}$$  But can somehow provide a explanation in the style of the quoted answer why this actually works? Paying off student loan or using that money for a downpayment on a house Identification of particles and anti-particles Transparent text cut out of background I've installed Ubuntu 16.04 on my Asus X555LJ.  But I don't know how to install drivers or how to enable function keys to controls screen backlight brightness. Android Min SDK Version vs. Target SDK Version This may be a silly question, and I think I have looked elsewhere to find the answer...  Might be a path issue, but when I open the command line and type from the C:\&gt;: cd D:\  I cannot get to the D drive.  Even if I type: cd D:\&lt;folder name&gt;  The command.exe will auto-complete the line with the tab key, so it knows where I'm at.  It just doesn't print to screen the result or actually get me there.  This problem exists for the network drives as well. Now, if I use the chdir (cd) command like this: chdir D: or cd d: I get the print out of the D:\ below the command but it still says I'm in the C:\. I feel like I'm missing something simple.