# How to do complicated problem without messy mind?

I have trouble when doing complicated problem, when I look at a problem with so much information.
(e.g. deal with some concrete example such as proving a 'ugly' space is homeomorphic to another 'ugly' space by explicitly writing down the homeomorphism, proving two 'ugly' ring are isomorphic. Problems that have so many 'detail' stuff.)

May I use some real problems to explain?
1. Let $G$ be a finite group with odd order, prove that for any $g\in G$ there exist a unique $x\in G$ s.t. $x^2=g$.
2.(a) Prove that for each $d|n$, number of elements of order d in $(\Bbb{Z}_n,+)=\phi(d)$, where $\phi(d)$ is the Euler's totient function.
(b)Let $G$ be a finite group, $\forall n\in \Bbb{N}$, there exists at most n elements $g\in G$ s.t. $g^n=e$, prove that $G$ is cyclic.

For the first problem, I know how to do it at a glance (define a map $\phi:G\to G$ by $\phi(g)=g^2$ $\forall g\in G$, prove that it is an injection, then it will automatically be a bijection since G is finite. Condition 'odd order' will show up when proving injectivity), but one of my clever friends (I am probably the most stupid one among us) can not solve it even he was thinking about it for 5 minutes. But the second, which is a exam problem, I can not solve it even I have tried for two days, but he can, he solved it in the exam.

I think it is worth talking about what is happened in my mind when I was trying to solve problem 2:

Well let me solve (a) first. What do we have? $\Bbb{Z}_n$, a number d divides n and a totient function. We want to know something about elements of order d. Hmm, what fact do I know about order? Cyclic subgroup generated by that element...Oh yes, we have learnt Lagrange's theorem, it may be helpful because the problem say that $d|n$...But I have no idea, maybe move to (b) first?

(b) is talking about cyclic group and order of an element($g^n=e$), it must be related to (a). Well we can apply (a) to the condition $g^n=e$ and see how it goes. $g^n=e$ iff $ord(g)|n$, so let $ord(g)=d$, then...wait, there is no $\Bbb{Z}_n$. Oh, we can also let $n=|G|$ so that it may works together with Lagrange's theorem...(after a while) It doesn't work. How about let $n=ord(g)$? Wait maybe I can go back to (a) first...No idea, maybe back to (b)?...Or try (a) again?(Start getting lost)

In summary, when I do complicated problem, I jump between huge amount of information (or even parts of problem!) soon and start to do nothing but just mess everything up. My mind become messy. I can ponder tricky but 'clean' problem like problem 1, since I know where I should focus, although the approach may be strange and weird (but 'clean'), I can come up with suitable approach after spending some time on it. But when I do complicated problem like problem 2, I can just get lost in the sea of information and do nothing.

Does anyone has the same problem? How do you deal with it? Please give me some advice, thank you very much.

• Identify the essence of a problem and eliminate extraneous data. Jump to the proper conceptual level and not get lost in details. Focus on how to set up a problem, not on the computation needed to solve it. Practice. – David G. Stork May 5 '16 at 15:17
• What do you mean by 'set up' a problem? Could you please give a simple example? Thank you – Longitude May 5 '16 at 16:02
• "Set up" a problem means casting it into a mathematical form. Euler "set up" the Königsberg Bridge problem as a graph problem, for instance. – David G. Stork May 5 '16 at 18:01

For example ,in 2(a), I represent $n$ by a set of vertical bars.Each bar represents the largest power of a prime divisor of $n,$ so the set of bars represents the prime decomposition of $n.$ Now $d|n,$ so $n/d$ and $d$ are represented as sets of parts of these bars.
Then I saw that in order for $x$ to have order $d$ in $Z_n,$ the bars in its prime representation that are in common with the bars of $n$ must have the same heights as the bars of $n/d.$ So $(n/d)|x.$
And the rest of the bar-representation of $x ,$ that is, of $x/(n/d),$ must have prime factors (if any) not common to $n$ and therefore not common to $d .$ But since we want $1\leq x\leq n,$ and we have $x\geq n/d,$ this limits our possible choices for $x/(n/d)$ to values from $1$ to $d$ that are co-prime to $d.$