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40 votes

Conjectures for finite groups that fail with large counterexamples

According to this answer by Chain Markov on the post I linked to, here are a few, ordered from smallest to largest (known) counterexample: Automorphism group of a non-abelian group is non-abelian: ...
Robin's user avatar
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15 votes
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Test if a finite group is a symmetric group algorithmically

The goal should be to do it in time $O(n!^2)$, which is the size of the input (the multiplication table). First determine the partition of $G$ into conjugacy classes. This can be done efficiently with ...
Sean Eberhard's user avatar
14 votes
Accepted

A modification of a cyclic group that seems to break being a group. What is it?

I will let $a\oplus b $ denote $N-\text{mod}(a+b,N)$. This operation is not even associative, as $$(s\oplus0)\oplus0=s\ne N-s=s\oplus0=s\oplus(0\oplus0)$$ Actually, it is a commutative quasigroup, ...
Ricky's user avatar
  • 3,156
13 votes
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Is it possible for a normal subgroup of a finite group have greater number of elements in the minimal generating set?

Let me generalize @ahulpke's example: Let $V$ be the additive group and $H$ be the multiplicative group of a finite field $\mathbb{F}_{p^n}$. Then $H$ acts on $V$ by multiplication and we can define $...
Brauer Suzuki's user avatar
11 votes
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Is there a finite abelian group which is not isomorphic to either the additive or multiplicative group of a field?

The two results being used below are that the multiplicative group of a finite field is cyclic and the classification of finite abelian groups. For the former, see this MO thread for a discussion of ...
11 votes
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If $p> \sqrt{|G|}$ divides $|G|$, then $G$ contains a normal subgroup of order $p$

Let $H$ be a subgroup of order $p$ (Cauchy guarantees its existence , $p | |G||$) and assume $H$ is not normal. Then there exists a $g \in G$ with $H^g:=g^{-1}Hg \neq H$, whence (using Lagrange and $p$...
Nicky Hekster's user avatar
11 votes
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Can you completely determine a finitely presented finite group?

In fact, in this case you can determine the group! Finiteness is absolutely crucial here, however. Let's say our generators are $a_1,...,a_n$, our relators are $R_1,...,R_m$, and every element of our ...
Noah Schweber's user avatar
11 votes
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For which $n$ in $2048 < n < 4096$ is the exact number of groups of order $n$ up to isomorphism currently unknown?

tl;dr: the only "unknown" order is 3072. In the small groups library (available in the computer algebra systems GAP and Magma), many of the groups of order in the range from 2049 to 4095 are ...
Max Horn's user avatar
  • 1,862
11 votes
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How can I tell if a permutation can be expressed as the commutator of another 2 permutations?

For the question in the post (now edited): In any group other than the trivial group and the group of two elements, every element can be written as the product of two nontrivial elements. In ...
Arturo Magidin's user avatar
10 votes
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How does growth in the number of nonisomorphic groups tells us something about the number of ways subgroups can be put together?

I find this statement a bit vague. I guess it means something like if $N$ and $M$ are finite $p$-groups, then there is no absolute bound on the number of isomorphism classes of groups $G$ with a ...
Derek Holt's user avatar
  • 90.2k
10 votes

Is there a finite abelian group which is not isomorphic to either the additive or multiplicative group of a field?

The group $\mathbb{Z_2}\times\mathbb{Z_6}$ for example. It can't be the multiplicative group of any (necessary finite) field because it is not cyclic. And it can't be the additive group of any field ...
Mark's user avatar
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10 votes
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Are there $n>6$ such that there is exactly one non-abelian group of order $n$ up to isomorphism? This happens with $n=6$ with $S_3$

There are many more such numbers. They can be generated with the following piece of GAP code. ...
Martin Brandenburg's user avatar
10 votes

Conjectures for finite groups that fail with large counterexamples

Another answer from me, separate from my other answer since it was not on the list that I cited there: Suppose $G$ is the unique group of order $N$ (that is, $G$ is cyclic). Then $N$ has at most $2$ ...
Robin's user avatar
  • 3,247
10 votes
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If $m\mid n$ can we find a group $G$ of order $n$ with a subgroup $H$ of order $m$, i.e. can every divisibility be proved group theoretically?

More precisely, if $m\mid n$, can we always find a group $G$ of order $n$ and a subgroup $H$ of $G$ of order $m$? Yes. For each $n$, there exists a cyclic group of order $n$. In the cyclic group of ...
Shaun's user avatar
  • 45.2k
9 votes

A contradiction to Lagrange's theorem?

$(\mathbb Z_3,\times)$ is not a group, so the theorem does not apply. $(\mathbb Z_3 \backslash \{0\},\times)$ is indeed a group, but it cannot be a subgroup of $(\mathbb Z_3,\times)$ because, again, ...
SolubleFish's user avatar
  • 7,918
9 votes
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Elements of $\mathbb{Z}_6 / \mathbb{Z}_2$

You have to ask yourself: how do we view $\Bbb Z_2$ as a subgroup of $\Bbb Z_6$? The only possible way to do this is to identify $[0]$ with $[0]$ and to identify $[1]\in\Bbb Z_2$ with $[3]\in\Bbb Z_6$....
FShrike's user avatar
  • 40.7k
9 votes

In a finite group $G$, prove that if $a^2 = b^2 \neq e$, then $a = b$.

I don't believe it can be extended to even order because of the following counterexample: Consider the cyclic group $G=C_4=\langle g\rangle$. Taking $a=g,b=g^3$ gives $a^2=g^2=b^2$, but $a\neq b$
Yaneda's user avatar
  • 702
9 votes
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Is this set, defined in terms of conjugation, closed under inverses?

I was immediately convinced that the answer to this question had to be no, because there is no conceivable reason why it should be yes. So then it just became a matter of systematically searching for ...
Derek Holt's user avatar
  • 90.2k
9 votes

Test if a finite group is a symmetric group algorithmically

By the Coxeter presentation of the symmetric group, a homomorphism $S_n \to G$ can be described equivalently as a sequence of elements $s_1,\dotsc,s_{n-1} \in G$ such that $$s_i^2 = 1,\, (s_i s_{i+1})^...
Martin Brandenburg's user avatar
9 votes
Accepted

Let $G$ be a finite group and $f$ a non-trivial automorphism of $G$ such that for each $x\in G, f(x)=x$ or $f(x)=x^{-1}$. Prove that $G$ is solvable.

Let $H = \{ x \in G \mid f(x)=x \}$. Then $H$ is a subgroup of $G$ and, since $f$ is nontrivial, there exists $g \in G \setminus H$. Then for all $h \in H$, $gh \not\in H$, and so $f(gh) = g^{-1}h= (...
Derek Holt's user avatar
  • 90.2k
8 votes

Is there a finite abelian group which is not isomorphic to either the additive or multiplicative group of a field?

There is a slightly easier argument, which does not require knowledge of the groups of finite fields. Any group of order $14$ cannot be the additive or multiplicative group of a field since neither $...
Brauer Suzuki's user avatar
8 votes
Accepted

Artin's theorem for the linear representation of finite groups

The Wikipedia page on Artin's theorem, https://en.wikipedia.org/wiki/Artin%27s_theorem_on_induced_characters, tells you the actual theorem Artin cared about proving: every character on a finite group $...
KCd's user avatar
  • 46.2k
8 votes

What exactly is the orbit-stabilizer theorem?

The way I state it in my final year group theory course is: Let $G$ act on $\Omega$ and let $\alpha \in \Omega$. Then there is a bijection between the right cosets $G_\alpha g$ of $G_\alpha$ in $G$ ...
Derek Holt's user avatar
  • 90.2k
8 votes

Growth rate of orders of perfect groups

This doesn't answer the question that you asked, but if we let ${\rm Perf}(\le n)$ be the number of isomorphism classes of perfect groups of order up to $n$, then we have the bounds $$n^{l(n)^2/108-cl(...
Derek Holt's user avatar
  • 90.2k
8 votes
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Center of group of order 3773

Let $P$ be a $7$-Sylow subgroup, $Q$ is an $11$-Sylow subgroup. Then $P,Q$ are normal in $G$, $P\cap Q=\{e\}$ and by comparing cardinalities, $G=PQ$. It is well known that in this case we have: $G\...
Mark's user avatar
  • 40.4k
7 votes

Maximal subgroup of a finite group contains all other subgroups.

I want to prove that all other proper subgroups of $G$ except $M$ are contained in $M$. No, it is false. Counter-example, Klein 4 group. $G=\{e, a,b,c \}$, where $\{e,a \}$, $\{e,b\}$, $\{e,c\}$ are ...
MathFail's user avatar
  • 21.1k
7 votes
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Finite non abelian simple groups whose order is not divisible by 8

There are infinitely many: the groups ${\rm PSL}(2,q)$ with $q \equiv \pm 3 \bmod 8$.
Derek Holt's user avatar
  • 90.2k
7 votes
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Understand proof of lemma about existence/uniqueness of group generated by a set

The definition says that the subgroup generated by the subset $S$ is a subgroup with certain properties. But just making that statement does not guarantee that such a subgroup exists. So really before ...
Derek Holt's user avatar
  • 90.2k
7 votes
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Does $S_n$ always embed into $GL_{n-1} (\mathbb{F}_p$)?

Okay, I was being a little flippant. There are actually some nontrivial details to check here because of positive characteristic issues so let's check them. Let $F$ be any field. $S_n$ acts faithfully ...
Qiaochu Yuan's user avatar
7 votes

When are all normal subgroups of a direct product of finite groups a direct product of normal subgroups?

Edit. Okay, found a simpler-to-state condition; I'm keeping the original response below, between the two horizontal lines so that (i) it gives context; and (ii) so the folks who upvoted don't feel ...
Arturo Magidin's user avatar

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