Prove there is no homomorphism from $Z_{16} \oplus Z_2 $ ont $Z_4 \oplus Z_4$ Prove there is no homomorphism from $Z_{16} \oplus Z_2 $ onto $Z_4 \oplus Z_4$   I have no idea at all how to attempt such question. I have read some solutions in which we find the cardinality of $kerm\phi$ for some $\phi mapping $ that is onto using the first theorem of homomorphism.  After that I don't understand anything  Is there a general approach to these questions ?
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.
99% of the time, I'm clear on when I should use "a" versus "an." There's one case, though, where people &amp; references I respect disagree.  Which of the following would you precede with "a" or "an," and why?   FAQ FUBAR SCUBA   [Note: I've read the questions  and , but the rules given there don't necessarily apply here.]    [Edited to add]  Here's a shorter (and hopefully clearer) version of the question…  In written English, which is correct (and why): "a FAQ" or "an FAQ"?  Some references with differing opinions:   an: the  and  a: the Microsoft Manual of Style for Tech Publications, 3e either: the  and How to mass rename files with ill-formed numbering? I use a Canon EOS Rebel T3i (600D) with the kit lens. I have experimented a lot with different settings and modes, but I feel like I have wasted many good shots that could have looked amazing, if only they were a little bit sharper.  I use auto-focus on most of my pictures and make sure to choose the right focus point, but all of my photos are still a little bit out of focus. When I zoom in, I can see that the pixels are not very crisp or well-aligned.  I am not sure if this issue is caused by my equipment or skills. Error when clipping raster in QGIS: "'ascii' codec can't encode character"? I have done it many times (clipping a raster with a polygon layer or by extension) but today I got an error for the first time and I don't know how to solve it. I am always using the same type of raster layers. When running the clipping tool I got the following message:  gdal_translate -of GTiff -ot Byte -projwin 410282.185971 4694950.41162 410783.882475 4694524.96815 -co COMPRESS=DEFLATE -co PREDICTOR=1 -co ZLEVEL=6 -co TFW=YES "C:/Users/FERRAN ALA/Desktop/Justificació Fontanals/QGIS/Topo 1_5000/bt5mv20sd0f283077st1r031.sid" "C:/Users/FERRAN ALA/Desktop/Justificació Fontanals/QGIS/background/image.tif"   GDAL command output:     'ascii' codec can't encode character u'\xf3' in position 201: ordinal not in range(128) See log for more details   Does anybody know what that is? I am a new user for Ubuntu. I downloaded 13.10 yesterday and made a bootable USB with universal USB installer on my Lenovo IdeaPad Z580 laptop which is now using UEFI; everything appeared OK.  When I booted from the LiveUSB I got the choices of trying or installing Ubuntu but both of them keep giving me these error:  [ 1.929082] kernel panic-not syncing vfs unable to mount root fs on unknown-block (0.0)   Please tell me what is going wrong? Why does $x$ divided by zero not equal $x$? After all, $x$ is not being divided by anything. $\dim(U_1+U_2) = \dim U_1 +\dim U_2 - \dim(U_1\cap U_2).$  I want to make sure that my intuition is correct. Suppose we have two planes $U_1,U_2$ though the origin in $\mathbb{R^3}$. Since the planes meet at the origin, they also intersect, which in this case is a one-dimensional line in $\mathbb{R^3}$. To obtain the dimension of $U_1$ and $U_2$, we add the dimensions of the planes (4), and the subtract the dimensions of the line (1), which results in (3).   *additional question(s):   Can we generalize this notion to $\mathbb{F^{n}}$?  Suppose we have an additional case where $U_1$ and $U_2$ are planes in $\mathbb{R^3}$, but $U_1 \subseteq U_2$. In this instance, $dim(U_1 + U_2) &lt; 3$, because the first two-dimensional plane is contained in the second and as a result, the dimensions of the subspaces when summed cannot exceed two. Since both subspaces $U_1,U_2$ are two dimensional and $U_1 \subseteq U_2$, then their intersection is also two-dimensional, concluding $dim(U_1+U_2)=2+2-2 = 2$.  Is this proper intuition? Prove that if $\gcd(a,b)=1$ then $\gcd(a^m, b)=1$ "I am going to have a walk" My teacher said "I am going to have a walk" today in class and to me it seems incorrect. Correct variants that I know are:    Going for a walk Going to take a walk Going to walk  ...   Is this correct? If so, why? In single-blind peer-review, can you reveal your identity without the editor's consent? In my field (chemistry), review is always done in a single blind process, i.e. the author does not know who the referees were, but the referees do know who the authors are. In ten years, I have never seen a referee breaching this anonymity, e.g. by signing their review, unless they got prior approval by the editor (in order to continue discuss things further with the authors, once the manuscript was accepted for publication).  So, it seemed logical to assume that in single blind peer-review, the reviewer should not disclose his identity without the editor's consent. Yet, some people have told me ( and ; also IRL a friend from human sciences) that they have seen people sign their reviews, or write emails to them after the review but before the paper is published.  So, in single-blind peer-review, can you reveal your identity without the editor's consent? Does it depends on the customs of each field, or is there a hard rule? (in which case, the few anecdotes I heard were outliers) What does "Incompatible types: void cannot be converted to ..." mean? Bash-like, non-rotating filename tab completion with PowerShell Under Windows, using PowerShell, is it possible to make the filename suggestion (using the Tab key) work like with bash under Linux?  Let's assume a directory contains 3 directories: example1, example and examination.  On Windows (by default), if you type cd ex and Tab, it will auto-suggest cd .\examination. Subsequent uses of Tab will rotate the suggestion between example1, example and examination.  Using bash (typically under Linux), if you type cd ex and Tab, it will auto-complete until the character where the filenames start to differ: cd exam. Another use of Tab then suggests which directories are available:  $ cd exam examination/ example1/    example2/   Then, if you type p and Tab, it will auto-complete to example, and so on (here, offering a choice between example1 and example2).  Is there a way to make the filename suggestion work in a tree-like search like this, instead of rotating through the names?  (I'm mainly interested in file names. Bash can also refine this selection based on the file type, for example only directories for cd, which would be good, but having the file name selection itself would be enough.) Idea behind proof that $\frac{21n+4}{14n+3}$ is irreducible for all $n$ Shape of the universe? Is the convex hull of closed set in $\mathbb R^{n}$ is closed? Is convex hull of closed set in $\mathbb R^{n}$ closed? Is the set of polynomials of degree less than or equal to $n$ closed? This question is in relation to the space $C(I)$, $I = [a, b]$. Define $P_n =\{ a_0+\dots+a_nx^n \mid a_i \in \mathbb{R}\}$ (any or all $a_i$ could be zero); clearly $P_n \subset C(I)$. The norm I'm using is $\lVert f\rVert_I = \sup_I |f(x)|$. Is $P_n$ closed under $\lVert\cdot\rVert_I$?   I am almost sure the answer is "yes", but I can't seem to prove it. My first instinct was to biject  $P_n$ to $\mathbb{R}^{n+1}$, using coefficients as coordinates, and prove that sequences of degree-$n$ polynomials converge to degree-$n$ polynomials, but I can't prove that the metric $\lVert\cdot\rVert_I$ is equivalent to the standard metric on $\mathbb{R}^{n+1}$. Next, intuitively given a function $f \in C(I), \notin P_n$ I should be able to find some $\epsilon &gt; 0$ such that there are no low-degree polynomials "nearby", then that function was not a limit point of $P_n$ so $P_n$ is closed. Again I have no idea how to prove this. What should I do? Currently using Ubuntu 18.04, and my wireless driver is not being detected.  I've included a .  $ dmesg | grep iwl [    6.212724] Loading modules backported from iwlwifi [    6.212724] iwlwifi-stack-public:master:6988:0d8892f2 [    6.224583] iwlwifi 0000:00:14.3: enabling device (0000 -&gt; 0002) [    6.229811] iwlwifi 0000:00:14.3: Direct firmware load for iwl-dbg-cfg.ini failed with error -2 [    6.230540] iwlwifi 0000:00:14.3: Direct firmware load for iwlwifi-9000-pu-b0-jf-b0-39.ucode failed with error -2 [    6.230547] iwlwifi 0000:00:14.3: Direct firmware load for iwlwifi-9000-pu-b0-jf-b0-38.ucode failed with error -2 [    6.230552] iwlwifi 0000:00:14.3: Direct firmware load for iwlwifi-9000-pu-b0-jf-b0-37.ucode failed with error -2 [    6.230557] iwlwifi 0000:00:14.3: Direct firmware load for iwlwifi-9000-pu-b0-jf-b0-36.ucode failed with error -2 [    6.230562] iwlwifi 0000:00:14.3: Direct firmware load for iwlwifi-9000-pu-b0-jf-b0-35.ucode failed with error -2 [    6.236670] iwlwifi 0000:00:14.3: loaded firmware version 34.0.0 op_mode iwlmvm [    6.248314] iwlwifi 0000:00:14.3: Detected Intel(R) Dual Band Wireless AC 9560, REV=0x318 [    7.264081] iwlwifi 0000:00:14.3: SecBoot CPU1 Status: 0x3, CPU2 Status: 0x2477 [    7.264084] iwlwifi 0000:00:14.3: Failed to start INIT ucode: -110 [    7.276165] iwlwifi 0000:00:14.3: Failed to run INIT ucode: -110   sudo modprobe iwlwifi returns no output. A riddle about guessing hat colours (which is not among the commonly known ones) This is a riddle I heard recently, and my question is if someone happens to know the solution. I'm asking this out of curiosity more than anything else.  So here it is. The riddle is one of the countless variations of the "prisoners have to guess their hat colour" puzzle. $n$ prisoners are put a hat on top of their head, which can be red or blue. The colours are chosen at random by $n$ independent fair coin tosses. Then each prisoner can guess their own hat colour (red or blue) or pass. The prisoners can see each other, but not hear each other's calls and of course they have no other means of communication. This means that each call can only depend on the other prisoners' hat colours. However, before the distributing of hats begins, the prisoners are told the rules and can agree on a strategy. The prisoners win iff no prisoner guesses wrong and at least one prisoner guesses right. Which strategy should the prisoners use so that the winning probability becomes maximal?  Some remarks:   A simple strategy is that one player just guesses and all other players pass, so that the maximal probabilty is at least 1/2. For $n=2$ this strategy is optimal. For $n=3$, there is a strategy that wins in 6 out of 8 cases: When a player sees (red,red) he guesses blue, for (blue,blue) he guesses red, and otherwise he passes. More generally this shows that the maximal probability is at least 3/4 for $n\ge 3$. It's possible to show that any strategy fails for at least 2 hat colour configurations (unless $n=1$), which shows that the above strategy is optimal for $n=3$. For $n=4$ there are more than $10^{15}$ strategies, and for $n=5$ it's about $10^{38}$ strategies, making it quite infeasible to just use a brute-force computer program (maybe for $n=4$ it's possible when exploiting the obvious symmetries). When changing the rules slightly by forbidding players to pass, then the maximal winning probability is always 1/2. This is a nice little exercise.   Actually I heard the riddle only for $n=3$ and then thought about the general $n$. So it's entirely possible that there is no nice solution. Is there a way to detect the collapse in wave function of entangled particles instantaneously? Is there a machine or instrument which will notify us instantly when the wave-function has collapsed if we have access to only one of the entangled particles? Note that the wave function collapse will be triggered by observing the particle which we do not have. Or in other words is there a way to identify (instantaneously) if the entangled pair of a particle has been observed and the wave function has collapsed. Why does the area of a bar in a histogram represent the frequency of that particular class In a bar-graph, the height of the bars represent the frequency of a particular category but it is counter-intuitive that in a histogram, instead of the height of the bar(which represents the frequency density), the area of the bar represents the frequency of the particular class. Why is this so?