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Mar
17
comment How many nonabelian groups up to isomorphism are of the order $p^4q^4$?
Are you sure you really want to find all non-abelian groups of order $p^4q^4$, not just for the abelian ones? If yes, you could get acquainted with their $p$- rsp. $q$-Sylow subgroups by looking at the links given in the comments to these questions math.stackexchange.com/questions/1095008/… and math.stackexchange.com/questions/1162977/groups-of-order-p4 .
Mar
16
comment Group Permutations Proof
@JackM: Thanks, the question had already (for my taste too) many answers.
Mar
16
comment Prove: There is a $g \in G$ such that $\forall$ $x \in X: g \circ x \neq x$
This looks better! (And looking at Peter's last question he does know Burnside's lemma.)
Mar
16
comment Prove: There is a $g \in G$ such that $\forall$ $x \in X: g \circ x \neq x$
Why the upvote? Peter does not know Burnside's lemma (and what's the connection between having no fixed point free element and the fixed points of (which??) $g$ being the whole set? (I see no indication that the $g$ in the exponent is the identity.)).
Mar
16
comment Mapping vector spaces over two different fields?
Fields have a group structure given by addition and, if you take away $0$, one given by multiplication. Looks like you thought only of multiplication. What becomes out of $(a+b)v = av+bv$? (If you ask instead for a ring homomorphism, there will be only injective ones.)
Mar
16
comment Prove: There is a $g \in G$ such that $\forall$ $x \in X: g \circ x \neq x$
Typo: we can conclude that #Gx > 1 is correct.
Mar
16
comment Prove: There is a $g \in G$ such that $\forall$ $x \in X: g \circ x \neq x$
Try calculating the order of the set $\{(x,g)\in X\times G\mid g\circ x = x\}$ in two different ways: Once as sum over the elements of $G$ and once as sum over the elements of $X$.
Mar
16
comment Prove: There is a $g \in G$ such that $\forall$ $x \in X: g \circ x \neq x$
If you knew this lemma, you could use it nicely. I just wanted to know what tools you have at your disposal.
Mar
16
comment Group Permutations Proof
$a^6 = e$ tells you that the order of $a$ divides $6$. $ab = ba^2$ is better written as $b^{-1}ab = a^2$ telling you that $a$ is conjugated to $a^2$ via $b$, which implies that the order of $a$ cannot be divisible by $2$ (conjugated elements have the same order).
Mar
16
comment Prove: There is a $g \in G$ such that $\forall$ $x \in X: g \circ x \neq x$
Do you know en.wikipedia.org/wiki/Burnside%27s_lemma ?
Mar
15
comment computing the orbits for a group action
If you like to use a bit more of Galois theory instead, you could look also at the degrees of the minimal polynomials of all elements of $\mathbb{F}_9$ over $\mathbb{F}_3$. The Galois group acts transitively on roots of the minimal polynomials.
Mar
15
comment computing the orbits for a group action
Do you know the number of fixed points and the order of your group?
Mar
13
comment Computing the order of a group element
My first comment shows that my other comments are wrong: To factor $p=2qr+1$ just break the RSA with modulus $qr$. The order of elements in $\mathbb{Z}/qr\mathbb{Z}^\times$ give divisors of $\lambda(qr)$ that allow factoring $qr$.
Mar
13
comment Let $H,K$ be finite groups , if for any finite group $G$ , $h(G,K)=h(G,H)$ holds , then is it true that $i(G,H)=i(G,K)$ for any finite group $G$ ?
@SaunDev: So you can rephrase your question slightly simpler in the style of math.stackexchange.com/questions/1161733/…
Mar
12
comment $G$ is a simple group of order $60$.Then $G$ contains a subgroup of order 12
$n_2 = 3$ gives you a non-trivial homomorphism to $S_3$.
Mar
12
comment Can I recover a group by its homomorphisms?
@Turion: Five minutes wikipedia (en.wikipedia.org/wiki/Residually_finite_group) and you'll know enough about it (e.g., that you should have included this condition from the beginning in your question). It is as essential as "finitely generated": f.g. prevents that your numbers are infinite and r.f. prevents that they are all zero.
Mar
12
comment Computing the order of a group element
@DRF: We seem to agree. My comment with $p=2qr+1$ was about the fact, that you won't get lucky often enough.
Mar
12
comment Computing the order of a group element
@DerekHolt: What use does an order-finding algorithm have if you can get only elements of order $p-1$ or $\frac{p-1}{2}$?
Mar
12
comment Computing the order of a group element
I somehow doubt that your memory is correct: If $p = 2qr+1$ where $q$ and $r$ are primes both of about the same size, then one is unlikely to find elements of order $q$ without knowing $r$. Maybe you read the implication the other way round (which is true)?
Mar
12
comment If G is a group such that all of its proper subgroups are abelian, then G itself must be abelian
$Q_8\times\mathbb 1\times 1\le Q_8\times\mathbb Z_2\times\mathbb Z(3^{\infty})$ is abelian?