My teacher told me to prove this using the proof by contradiction. So, I am going to assume that there's an infinite group with finitely many subgroups. Then this will lead to finding at least one element with an infinite order, from which I can derive a contradiction.   But I am lost how I can prove this...   I am thinking if the group has its subgroups, then every element in those subgroups must also have infinite order as they are in their original group, aren't they? So, my conclusion is that every element in infinite group's subgroups must have infinite order except identity element.  But how does this contradict the group is infinite?
I'm trying to prove that any group $G$ of infinite order has an infinite number of subgroups.  I think that if the group has an element of infinite order, then it's easy because I can take the groups generated by the powers of this element.  What if it doesn't? Every element generates a cyclic subgroup. Every element belongs to at least one cyclic subgroup (that generated by itself). So the group is the union of its cyclic subgroups. If all these are finite, we would have to have an infinite collection of subgroups anyway.  Is that correct?
What constitutes protons? When I see pictures, I can't understand. Protons are made of quarks, but some say that they are made of 99% empty space. Also, in this illustration from Wikipedia, what's between the quarks? What was the issue solved by the new "using" syntax for template typedefs? What's the best way of handling permissions for Apache 2's user www-data in /var/www? Has anyone got a nice solution for handling files in /var/www? We're running Name Based Virtual Hosts and the Apache 2 user is www-data.  We've got two regular users &amp; root. So when messing with files in /var/www, rather than having to...  chown -R www-data:www-data   ...all the time, what's a good way of handling this?  Supplementary question: How hardcore do you then go on permissions?  This one has always been a problem in collaborative development environments. Do these new insights into prime numbers affect encryption security? reports:  Two mathematicians have uncovered a simple, previously unnoticed property of prime numbers [...]. Prime numbers, it seems, have decided preferences about the final digits of the primes that immediately follow them. Among the first billion prime numbers, for instance, a prime ending in 9 is almost 65 percent more likely to be followed by a prime ending in 1 than another prime ending in 9.  Do these insights make breaking keys based on primes more mathematically plausible in a shorter amount of time? How to decide on the correct number of clusters? We find the cluster centers and assign points to k different cluster bins in  which is a very well known algorithm and is found almost in every machine learning package on the net. But the missing and most important part in my opinion is the choice of a correct k. What is the best value for it? And, what is meant by best?  I use MATLAB for scientific computing where looking at silhouette plots is given as a way to decide on k . However, I would be more interested in Bayesian approaches. Any suggestions are appreciated. How to calculate polygon areas and perimeters using QGIS? Points at integer distance How many points can one can place in $\mathbb{R}^n$, with the requirement that no $n+1$ points lie in the same $\mathbb{R}^{n-1}$-plane, and the euclidean distance between every two points is an integer? $R/Ra$ is an injective module over itself Let $R$ be a PID, $a\in R$ be a nonzero nonunit in $R$. Prove that $R/Ra$ is an injective module over itself.   If $R$ is a PID, every $R$- divisible module is injective, but the question concerns with $R/Ra$-module, so I have no idea to solve this problem. Help me. Some spells, such as Forbiddance, say that under certain circumstances, they last "Until Dispelled".  Other spells, such as Mordenkainen's Private Sanctum, say that under certain circumstances, they become "Permanent".  What is the difference between these two terms?  Does "Permanent" mean that the effect can never be dispelled? Best way to soft brute-force your own GPG/PGP passphrase? Failed to connect to socket /com/ubuntu/upstart: Connection refused: Errors were encountered while processing: runit Single-word synonym for "immensely valuable"? I am looking for some (may be rarely know to nonnative English speakers) synonyms for  "immensely valuable" or "extremely valuable". Inspired by question :  When do I need to invoke \phantomsection?  In what condition is invoking \phantomsection necessary? wrapfig: frame around image and caption Does *first of all* always go with *second of all*? According to , first of all is twice as popular as firstly, but secondly is a hundred times more popular than second of all.  When I present two points, do I have to be consistent in the choice of these phrases (if I use firstly, should the other be secondly)?  Because If I choose the most popular, my sentence would be like "First of all he is rich, secondly he is also smart." Which exhibits kind of inconsistency. How does it sound? How do I escape the wildcard/asterisk character in bash? For example:  me$ FOO="BAR * BAR" me$ echo $FOO BAR file1 file2 file3 file4 BAR   and using the \ escape character:  me$ FOO="BAR \* BAR" me$ echo $FOO BAR \* BAR   I'm obviously doing something stupid.  How do I get the output BAR * BAR? Here is the top output which I gathered:    I noticed that top is showing vlc's cpu usage is > 100%.  Can anybody please justify me why it is showing like this, whether this is a bug in top application or something else. The IPA symbol for the labiodental flap (ⱱ) Say for a game with irregular hotspots (say Risk, or a strategy game like Paradox's Hearts of Iron or Crusader Kings), how do you   Define a non-rectangular hotspot of irregular shape (much like a territory on a map?), in code? What tool can you use to trace out the hotspot and import into your game? What is the algorithm for detecting if someone has clicked within such a hotspot? Problem:  Let $A_{n.n}$ be square complex matrix. Prove the following: $$\left \| A \right \|=\left \| A \right \|_{HS}\Leftrightarrow  rank(A)\leqslant 1$$. Where $\left \| . \right \|_{HS} $ is the Hilbert Schmidt Norm.  Please read my solution and tell me whether it is correct. If not, let me know where the mistake is.  Proof of the implication $\Leftarrow $:   If $rank(A)=0$, then in this case $A=0$. It follows that $\left \| A \right \|=\left \| A \right \|_{HS}=0$ If $rank(A)=1$, than: $ A=\begin{pmatrix} A_{1}\\  \alpha _{2}A_{1}\\  ...\\  \alpha _{n}A_{1} \end{pmatrix}$ where $A_{1}$ is the first row of $A$   and in this case: $\left \| A \right \|=\left \| A \right \|_{HS}=\left \| A_{1} \right \|\sqrt{1+\alpha _{1}^{2}+...+\alpha _{n}^{2}}$  Proof of the implication $\Rightarrow $   We know that $\left \| A \right \|=max_{\left \| x \right \|=1}\left \| Ax \right \|$  On the other hand: $ A=\begin{pmatrix} A_{1}\\  A_{2}\\  ... \\ A_{n}  \end{pmatrix}$. So, $$\left \| Ax \right \|=\left \| \begin{pmatrix} \left \langle A_{1},x \right \rangle\\  \left \langle A_{2},x \right \rangle\\  ...\\   \left \langle A_{n},x \right \rangle\end{pmatrix} \right \|=\sqrt{\left \langle A_{1},x \right \rangle^{2}+\left \langle A_{2},x \right \rangle^{2}+...+\left \langle A_{n},x \right \rangle^{2}} $$  Where $ \left \langle A_{i},x \right \rangle$ is the inner product of $A_{i}$ and $x$  Using the Cauchy-Schwarz inequality: $\left \langle A_{i},x \right \rangle^{2}\leq \left \| A_{i} \right \|^{2}\left \| x \right \|^{2}$, we get: $\left \| A \right \|=max_{\left \| x \right \|=1}\left \| Ax \right \|\leq max_{\left \| x \right \|=1}\left \| x \right \|\left \| A \right \|_{HS}=\left \| A \right \|_{HS}$. In order to have the equality: $\left \| A \right \|=\left \| A \right \|_{HS}$, we should have: $\left \langle A_{i},x \right \rangle^{2}=\left \| A_{i} \right \|^{2}\left \| x \right \|^{2} $ which occurs only if $A_{i}=\lambda _{i} x$. So, the rows of A are dependent, which implies that $rank(A)=1$. Note that the case $rank(A)=0$ happens when $A=0$  Please let me know if my solution makes sense.