Wikipedia:Reference desk/Archives/Mathematics/2013 September 2

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September 2

order of intersection of cyclic subgroups

Hi,

I recently came across the claim that if $G$ is a cyclic group and $A,B\subset G$ are subgroups with orders $a,b$ respectively, then the order of $A\cap B$ is $gcd(a,b)$.

I first tried to prove that $A\cap B$ was cyclic in the hope that it would make things easier. This wasn't too difficult. I first proved that any subgroup of a cyclic group is cyclic, then the fact followed since $A\cap B$, being the intersection of two subgroups is itself a subgroup.

I then showed that if $G$ is generated by $g$, and n,m are the smallest positive integers such that $g^{n},g^{m}$ generate $A$ and $B$ respectively then the smallest power of $g$ that generates $A\cap B$ was equal to the lowest common multiple of $m$ and $n$, but this approach has got me nowhere.

Help please?

Neuroxic (talk) 11:20, 2 September 2013 (UTC)

$|G|=k\cdot (a/\gcd(a,b))\cdot (b/\gcd(a,b))\cdot \gcd(a,b)$, for some integer $k$. So $n=kb/\gcd(a,b)$, and $m=ka/\gcd(a,b)$.

So $lcm(n,m)=k\cdot lcm(b/\gcd(a,b),a/\gcd(a,b))$. But these are relatively prime, so $lcm(n,m)=k\cdot (a/\gcd(a,b))\cdot (b/\gcd(a,b))$. So $lcm(n,m)\cdot \gcd(a,b)=|G|$. From this it follows that $g^{lcm(n,m)}$ generates a group of order $\gcd(a,b)$.--80.109.106.49 (talk) 12:45, 2 September 2013 (UTC)

Minimizing the largest eigenvalue

Given

$a\in \mathbb {R} ^{n}$,

$b\in \mathbb {R}$,

$A_{0},\cdots ,A_{n}\in \mathbb {S} ^{n\times n}$.

Show that

$\min _{x\in \mathbb {R} ^{n}}\sup _{u\in \mathbb {R} ^{n}}{\frac {u^{T}(A_{0}+x_{1}A_{1}+\cdots +x_{n}A_{n})u}{(a^{T}x+b)u^{T}u}}=\min _{x\in \mathbb {R} ^{n}}\sup _{u\in \mathbb {R} ^{n}}{\frac {u^{T}({\frac {1-a^{T}x}{b}}A_{0}+x_{1}A_{1}+\cdots +x_{n}A_{n})u}{u^{T}u}}$

(assuming I've done my preliminary calculations correctly)

AnalysisAlgebra (talk) 16:01, 2 September 2013 (UTC)