Linked Questions

41
votes
3answers
8k views

$|G|>2$ implies $G$ has non trivial automorphism

Well, this is an exercise problem from Herstein which sounds difficult: How does one prove that if $|G|>2$, then $G$ has non-trivial automorphism? The only thing I know which connects a group ...
38
votes
5answers
3k views

Advantage of accepting the axiom of choice

What is the advantage of accepting the axiom of choice over other axioms (for e.g. axiom of determinacy)? It seems that there is no clear reason to prefer over other axioms.. Thanks for help.
48
votes
6answers
2k views

Is there a non-commutative ring with a trivial automorphism group?

This question is related to this one. In that question, it is stated that nilpotent elements of a non-commutative ring with no non-trivial ring automorphisms form an ideal. Ted asks in the comment for ...
15
votes
5answers
2k views

Fields of arbitrary cardinality

So I took an introductory abstract algebra course a few semesters ago, and we were shown that groups and rings can both be made into products, i.e. if I have some group $G$ (resp. ring $R$) and some ...
19
votes
1answer
2k views

Proving “every set can be totally ordered” without using Axiom of Choice

It is known that the statement "every set can be totally ordered" is strictly weaker than Axiom of choice. How does one go about proving without using AC?
3
votes
3answers
2k views

If $ G $ has no non-trivial automorphism, then $ G $ is abelian and $ g^2 = e $ for all $ g \in G $ . [duplicate]

If $ G $ has no non-trivial automorphism, then $ G $ is abelian and $ g^2 = e $ for all $ g \in G $ . With the assumption, I dont know how to start the proof. If there is no non-trivial automorphism,...
4
votes
3answers
1k views

Centralizer of $Inn(G)$ in $Aut(G)$

Can the centralizer of Inn(G) in Aut(G), where G is preferably any non-abelian finite one, equal to Inn(G) itself? Clearly, such centralizer contains all $f$ in Aut(G) where $f(g)g^{-1}$ are in Z(G).
7
votes
1answer
916 views

Cyclic Automorphism group

Show that no group can have its automorphism group cyclic of odd order. I have shown it only if $G$ is cyclic, but I could not do that if $G$ is not cyclic. Can you help?
1
vote
1answer
286 views

Does there exist an abelian $2$-group of finite exponent that is not a direct sum of cyclic groups?

Does there exist an abelian $2$-group (an abelian group, all of whose elements have order a power $2$) of finite exponent that is not isomorphic to a direct sum of $2$-cyclic groups? The exponent of ...
11
votes
1answer
129 views

Choosing elements of linear orders

Is it consistent with ZF that there can be a countable family of linear orders, each isomorphic to $\mathbb Z$ (that is, every element has a unique predecessor and successor, and any two elements have ...
1
vote
2answers
174 views

Developing intuition for a world without AC

So after 25 years without doing any serious math, I've gotten the bug again. In my spare time (I have a full-time job as a lawyer), I've been starting to work my way through Set Theory: An ...
4
votes
1answer
173 views

Automorphisms of abelian groups and Choice

The latest question to be asked at the Group Pub Forum is a classic: can every group be realised as the automorphism group of a group? The answer is no, and the canonical answer is the infinite cyclic ...
2
votes
2answers
102 views

Proving that ${\rm Aut}(G)=\{{\rm id}\} \implies |G| \in \{1,2\}$. [duplicate]

The exercise asks to prove that if $G$ is any group with ${\rm Aut}(G) = \{{\rm id}\}$, then $g^2=1$ for all $g$ in $G$, $G$ is abelian, and if $G$ is finite, we'll have $|G| = 1$ or $2$. I managed ...
2
votes
1answer
141 views

Infinite dimensional vector spaces and inductive sets

Consider an infinite dimensional vector space $E$ and define $$S:=\left\{F \subset E\biggr| F\ne 0\text{ is a subspace of }E\right\}.$$ Endow $S$ with the reverse inclusion. Is it possible to find a ...
1
vote
1answer
111 views

One more question about eigenvalues in infinite dimension

Consider the following statement: If $U$ is a subspace of $V$ that is invariant under every operator on $V$ then $U=\{0\}$ or $U=V$. I can prove it easily assuming that $dim(V) = n$ is finite: If $...

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