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location From the Bay to LA
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visits member for 2 years, 1 month
seen 53 mins ago

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


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 $A_4$.
Aug
26
answered If $G$ is a simple $f$ an homomorphism, and $A\lhd H$ is such that $[H:A]=2$, show $f(G) \subset A$
Aug
25
comment Does $G$ always have a subgroup isomorphic to $G/N$?
@JorgeFernández Yes, take a look at this question.
Aug
25
answered Does $G$ always have a subgroup isomorphic to $G/N$?
Aug
25
comment Is any submanifold of $\mathbb{R}^{n}$ the zero set of some polynomial?
@QuangHoang Yes, open sets are (embedded) submanifolds.
Aug
20
comment Geometric interpretation of complex path integral
Nice question, +1.
Aug
20
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.
Aug
19
revised What does $\frac12(D_{2p}\times D_{2p})$ mean in group theory?
edited title
Aug
17
revised The spectrum of a commutative ring with unity and its “topology”
edited tags
Aug
16
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
Aug
16
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.
Aug
16
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 $
Aug
12
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.
Aug
11
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.
Aug
9
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?
Aug
9
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.
Aug
4
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 $1$.
Aug
2
awarded  Yearling
Aug
1
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 $K=F$.
Jul
23
revised How to prove “a group $G$ of order $72$ can't be a simple group”?
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