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 May12 comment Quadratic solutions puzzle (-1/2, -1/2) is a solution, but it is also said in the constraints that $a \neq b$. May11 comment Confused about the prime elements of a ring. @BillDubuque: Ok, thanks! It is just hard to wrap my head around. A lot of things in abstract algebra are requiring some serious rewiring of my brain. sigh May11 comment Confused about the prime elements of a ring. @BillDubuque: Ok, so then we can construct any numerator as a unique combination of primes (well known property of primes) and choose a specific bottom element as a particular unit so this would imply that R is a UFD then? May11 comment Confused about the prime elements of a ring. @BillDubuque: When you say $2$ is a unit, are you referring to the $p \times 2^{-1}$ is unnecessary just as $-1$ is unnecessary in prime factorization? May10 comment Need help determining what this center is isomorphic to Oh, whoops my bad, but $Z$ should have order $2$. That was me slipping up there, I didn't mean to write that, just mixed things around sorry about that. Thanks! (Modified original comment.) May10 comment Need help determining what this center is isomorphic to Ok, so I think the problem is I didn't know about p-group. In this case the argument should go: Since it is non-trivial, we know that the order is $2$, $4$ or $8$, but G is not abelian so the order cannot be $8$ and $2$ is guaranteed cyclic hence $Z$ has order $2$ and $|G/Z| = 4$ May8 comment Help show the following isomorphism cannot exist. Ok, so I think I get it, I made a small typo in my question, but was able to follow through and get $v^2 - v + 1 = 0$. But still, $x^2 - x + 1 \not\in \langle g(x) \rangle$. Thanks! May8 comment Help show the following isomorphism cannot exist. I still applied simplifications as if it was $\mathbb{Z}_3$ if that is what you are wondering... May8 comment Help show the following isomorphism cannot exist. Oops, sorry about that. May5 comment a question about abstract algebra,the order of $\Bbb Z_{5}[x]/ (x^3+x+1)$ Beat me too it too. +1 May4 comment Finding an ideal such that $\mathbb{Z}[x]/I \cong \mathbb{Z}[i]$. If understand this correctly: $f$ is the evaluation homomorphism, if we restrict the domain of $f : \mathbb{Z}[x] \to \mathbb{Z}[i]$ the same still holds. However, the first isomorphism theorem implies that the quotient group is isomorphic to this subgroup. However, $f$ is also surjective, so the quotient group is isomorphic to $\mathbb{Z}[i]$? May1 comment Show that if $x ≡ 1 (\text{mod } λ)$, then $x^3 ≡ 1 (\text{mod } λ^3)$… @Ebearr: Sorry my bad. (Referring to deleted answer) I guess it seemed a little too simple. Haha. Should of thought of that. Mar28 comment Proving that a sequence is increasing Cleaner than what I had wrote. +1 :) Also, I had a small hole in my writing now that I think about it... Mar20 comment Can I use Burnside's Theorem? Or should I take a different approach for this proof? Thank you for the response. Just to make sure I am understanding this: $3|[G : G_x]$ since $|G|= 3^3$ And we know that $|G|/|G_x|$ must be an integer, resulting in either $3^3$, $3^2$ or $3$, hence $|X'| \equiv 32 \mod 3$ meaning $|X'|$ has a strong lower bound of at least $2$ (since 32 mod 3 = 2)? Mar19 comment How do you apply an element on the left of a permutation? @ZachGershkoff: I was testing some small cases for a larger problem. I do not believe I am allowed to discuss the question that motivated this question, but I believe there was probably a flaw in the original question. Mar19 comment How do you apply an element on the left of a permutation? Hmm, I was testing out small cases for a larger problem. After reviewing your answer and the comments, I believe the question that motivated this question has a flaw in it. I would post the original but, I do not believe I am allowed to do so. Thanks for the help! Mar18 comment What about my proof is “nonsense”? @MikeMiller: I was thinking, that $x = ah$ for some $a \in G$ and $h \in H (\leq G)$, by closure of multiplication it has to be a member of $G$? Does this not make that much sense? Mar17 comment When does function composition commute? I think the example is correct, but the math is wrong: $f(g(x)) = (x^2)^3 = g(f(x)) = (x^3)^2 = x^6$ Mar16 comment Help with proving property of Rubik's cube. @vadim123: Now that you say it, I can see it. There always exists a side that can be rotated four times for $C_2$. But you can do this to to get two other sides that don't touch $C_1$ and touch all 7 other cubicles. Thank you for the help! Mar14 comment How should I continue my proof of this cycle property? (And did I make a mistake?) Also, I would edit it (except it doesn't meet the minimum 6 chars needed), but I am fairly certain you mean $x(a_i)$ not $x(a_1)$.