# BDub

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bio website location From the Bay to LA age member for 2 years, 1 month seen 53 mins ago profile views 659

Spent some time around Berkeley, now spending some time around LA.

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 11h comment A4 has no subgroup of order 6 @user156441 It's a well known fact that for any $n\geq 3$, $A_n$ is generated by $3$-cycles (here is a proof). So if $H$ contained all $3$-cycles, it would be $A_4$. Aug26 answered If $G$ is a simple $f$ an homomorphism, and $A\lhd H$ is such that $[H:A]=2$, show $f(G) \subset A$ Aug25 comment Does $G$ always have a subgroup isomorphic to $G/N$? @JorgeFernández Yes, take a look at this question. Aug25 answered Does $G$ always have a subgroup isomorphic to $G/N$? Aug25 comment Is any submanifold of $\mathbb{R}^{n}$ the zero set of some polynomial? @QuangHoang Yes, open sets are (embedded) submanifolds. Aug20 comment Geometric interpretation of complex path integral Nice question, +1. Aug20 comment Irreduciblility of $x^3 + 9x + 6$ in $\mathbb{Q}[x]$ If you can't use Eisenstein, can you use the rational roots theorem? It's probably easiest since the polynomial is a cubic. Aug19 revised What does $\frac12(D_{2p}\times D_{2p})$ mean in group theory? edited title Aug17 revised The spectrum of a commutative ring with unity and its “topology” edited tags Aug16 revised Let $W_1$ and $W_2$ be subspaces of a finite dimensional inner product space space. Prove that $(W_1 \cap W_2)^\perp=W_1^\perp + W_2^\perp$ added 138 characters in body Aug16 comment Let $W_1$ and $W_2$ be subspaces of a finite dimensional inner product space space. Prove that $(W_1 \cap W_2)^\perp=W_1^\perp + W_2^\perp$ @the8thone No problem. Aug16 answered Let $W_1$ and $W_2$ be subspaces of a finite dimensional inner product space space. Prove that $(W_1 \cap W_2)^\perp=W_1^\perp + W_2^\perp$ Aug12 comment How to find a basis of an image of a linear transformation? A basis of the image is the columns in the original matrix which correspond to the pivot columns in the row reduced matrix. So presumably the first and second columns of your row reduced matrix are pivot columns, so the first two columns of your original matrix are a basis. There may be a sign error in the answer of your book for the $2$ in the second basis vector. That's just my suspicion, I haven't actually worked it out. Aug11 comment Let $n \in \mathbb {Z}$. If $n^2$ is even, then $n$ is even. From your work, $4k^2+4k+1=2(2k^2+2k)+1$, so has remainder $1$ upon division by $2$, so is odd. Aug9 answered For the summation $\sum\limits_{n=0}^\infty\frac{(-1)^{n+1}n!}{1*3*5*…(2n+1)}$ when performing the Ratio test, why is the $(-1)^{n+1}$ term removed? Aug9 comment For the summation $\sum\limits_{n=0}^\infty\frac{(-1)^{n+1}n!}{1*3*5*…(2n+1)}$ when performing the Ratio test, why is the $(-1)^{n+1}$ term removed? The ratio test takes the limit of the absolute value of the ratio of consecutive terms. Since $|(-1)^{n+1}|=1$, it's irrelevant. Aug4 comment Confusion regarding proof of a proposition in Field Theory (Dumb Question) The dimension of a field extension is generally not the number of elements which you've adjoined. For instance if you have $\mathbb{Q}(\sqrt{2})$, you've only adjoined one element, but $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$, not $1$. Aug2 awarded Yearling Aug1 comment Extensions of degree $1$. The degree $[F:K]$ is just the dimension of $F$ viewed as a $K$-vector space. But $K$ is a $1$ dimensional $K$-vector space over itself, so if $[F:K]=1$, then $F$ has a subspace $K$ of equal dimension over $K$, but from linear algebra that means $K=F$. Jul23 revised How to prove “a group $G$ of order $72$ can't be a simple group”? added 2 characters in body