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

= 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
 * $$\min_{x \in \mathbb{R}^n} \sup_{u \in \mathbb{R}^n} \frac{u^T(A_0+x_1A_1+\cdots+x_nA_n)u}{(a^Tx+b)u^Tu} = \min_{x \in \mathbb{R}^n} \sup_{u \in \mathbb{R}^n} \frac{u^T(\frac{1-a^Tx}{b}A_0+x_1A_1+\cdots+x_nA_n)u}{u^Tu}$$

(assuming I've done my preliminary calculations correctly)

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