Salman
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 Jun 12 awarded Yearling May 3 comment Give an example of a group G with a subgroup H and a prime p such that a Sylow p-subgroup of H is not a Sylow p-subgroup of G A Sylow $p$-subgroup has to be maximal. In your case, you might have $p \mid [G:H]$. Jan 9 awarded Citizen Patrol Jan 4 comment How to prove the both identity (matrix) For the first: multiply by $(I + AB)$ on the left and $(I + BA)$ on the right. Expand both sides. For the second, use the fact that $A(I + BA)^{-1}$ is just $(I + AB)^{-1}A$. Move the term over to the other side and the identity reduces to $(I + AB)^{-1}(I + AB) = I$. Dec 8 awarded Caucus Aug 26 comment How to solve $e^x=x$? wolframalpha.com/input/?i=complex+root+of+e%5Ex+-+x Aug 21 answered The “Circle” is a Vector Space? Aug 19 comment Keisler Order: Saturated Ultrapowers I've elaborated a little more, though it's been a while since I've done anything with ultrapowers and my notes are spotty, so I wouldn't trust myself to flesh out the details without creating an enormous mess. But I think what I've written out is (hopefully) correct, and gives some idea of this. Aug 19 revised Keisler Order: Saturated Ultrapowers added 902 characters in body Aug 18 comment Keisler Order: Saturated Ultrapowers Sorry; yes, you are exactly right - $f$ itself need not be multiplicative (as you pointed out, it's very easy for it not to be), but there is a refinement which is. I just fixed it, I hope it's correct now. Aug 18 revised Keisler Order: Saturated Ultrapowers added 63 characters in body Aug 18 revised Keisler Order: Saturated Ultrapowers added 538 characters in body Aug 18 revised Keisler Order: Saturated Ultrapowers added 21 characters in body Aug 17 answered Axiomatizability of finite Isomorphic Classes Aug 17 answered Keisler Order: Saturated Ultrapowers Jun 12 awarded Yearling May 26 comment first order logic models The answer will (obviously) be 'no'. To see this, suppose $\varphi$ were such a sentence and either induct on complexity or put it in some sort of normal form ('prenex normal form' seems to be the phrase used on Wikipedia). The analysis may be tedious, but should come with relative ease. Jan 5 comment Must vectors in $\mathbb{R}^n$ have their “tail” at origin? When you think of $\mathbb{R}^n$ geometrically as a vector space, you are considering each vector as a direction arrow with a tail at the origin. So $(0, 10) - (0, 9)$ is an arrow $(0, 1)$ with a tail at the origin. Vectors here are only directions; the arrow you would want to draw from the head of one vector to the head of the other is not actually a vector unless you translate it back to the origin. If you try to define a vector space structure on $\mathbb{R}^n$ using the geometric interpretation of addition above, you'll run into problems (e.g. with additive identity). Nov 21 answered is it possible for two fields with the same characteristic to not be isomorphic? Nov 6 revised Evaluate the limit without L'Hospital fixed texing