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0 votes

Isomorphism on a Sylow-$p$-subgroup of a group $G$.

A subgroup $H\le G$ such that for any automorphism $a$ of $G$ we have $a(H)=H$ is called characteristic. Sylow subgroups are in general not characteristic, as we often have more than one, which by a ...
1 vote

Suppose we have a homomorphism $f:G\to\Bbb Z$ for a finite group $G$. Prove that $f(g)=0$ for all $g$ in $G$

Suppose (towards a contradiction) that there exists $g\in G$ with $f(g) \neq 0$. Then, since $G$ is finite, $|G|=n<\infty$. Then the order of $g$ must divide $n$. In any case, $g^n = 1_G$, so $f(g^...
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1 vote

Suppose we have a homomorphism $f:G\to\Bbb Z$ for a finite group $G$. Prove that $f(g)=0$ for all $g$ in $G$

$\Bbb Z$ doesn't have any non-trivial finite subgroups: every non-trivial subgroup is infinite cyclic. But $G/\rm ker(f)\cong f(G)\le\Bbb Z$ is a subgroup. That's the homomorphic image of a group is ...
1 vote

Suppose we have a homomorphism $f:G\to\Bbb Z$ for a finite group $G$. Prove that $f(g)=0$ for all $g$ in $G$

Suppose for the sake of contradiction that there is some $g \in G$ for which $0 \neq f(g) \in \mathbb{Z}$. Next, consider the sequence: $$f(g^1), f(g^2), f(g^3), \ldots$$ for which no two terms are ...
4 votes

Suppose we have a homomorphism $f:G\to\Bbb Z$ for a finite group $G$. Prove that $f(g)=0$ for all $g$ in $G$

Hint: The order of $f(g)$ divides the order of $g$ for all $g\in G$. Note that $\Bbb Z$ is torsion-free.
  • 41.4k
1 vote

Finitely generated group has automorphism mapping between two elements of the same order?

Automorphisms send conjugacy classes to conjugacy classes, while preserving elements' order. In $S_4$, the only distinct conjugacy classes of elements of the same order are the class of the ...
  • 1,295
7 votes
Accepted

Finitely generated group has automorphism mapping between two elements of the same order?

Take $G=S_3\times\mathbb{Z_2}$. It contains an element of order $2$ which is in the center, and an element of order $2$ not in the center. So clearly there is no automorphism which maps one of these ...
  • 34.5k
3 votes

Finitely generated group has automorphism mapping between two elements of the same order?

In $S_4$ there is no automorphism that takes a two cycle like $(12)$ to a product of two cycles like $(12)(34)$. In general, $S_n$ has only inner automorphisms when $n \ne 2,6$ and those preserve ...
  • 85.3k
2 votes

How to prove that $(G, \cdot)$ structure is an Abelian group where $G$ is a set of matrices determined by a certain rule?

The limits $|x| < 1$ and the presence of $\sqrt{1-x^2}$ suggest a trig substitution of $x = \sin\theta$ or $x = \tanh\eta$. The latter case shows more promise, giving $$ A(\eta) = \begin{bmatrix}\...
0 votes
Accepted

How to prove that $(G, \cdot)$ structure is an Abelian group where $G$ is a set of matrices determined by a certain rule?

One easily checks that $$A(x)A(y)=A\left(\frac{x+y}{1+xy}\right),$$ from which $$f(A(x)A(y))=f(A(x))+f(A(y))$$ follows. Since $f:G\to\Bbb R$ is moreover bijective, this proves both that $G$ is an ...
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2 votes

For a finite abelian $G$, $f: G\to G$ defined by $f(g)=g^2$ is an isomorphism iff $|G|$ is odd

Your analysis is good. In summary, if $G$ is a (multiplicative) group, then the map $f\colon G\to G$, $f(x)=x^2$ is a homomorphism if and only if $G$ is abelian; surjective if and only if it is ...
  • 233k
0 votes

For a finite abelian $G$, $f: G\to G$ defined by $f(g)=g^2$ is an isomorphism iff $|G|$ is odd

$f$ is injective if and only if $kerf=\{x\in G;g^{2}=0\}=0$. It is clear that $kerf=0$ if and only if $G$ has no element of even order (Because suppose that $g\in G$ such that $g^{m}=0$ where $m$ is ...
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4 votes
Accepted

For a finite abelian $G$, $f: G\to G$ defined by $f(g)=g^2$ is an isomorphism iff $|G|$ is odd

Suppose $|G|$ is odd. Let $g\in{ker(f)}$, meaning that $g^2=1$. Thus $g=1$ by Lagrange, otherwise $g$ is an element of order 2 in a group of odd order. Now, suppose $|G|$ is even. Then, $G$ admits an ...
  • 398
-1 votes

Isomorphism on a Sylow-$p$-subgroup of a group $G$.

What is true is the following. Some terminology: an automorphism of $G$ is said to be fixed-point free, if it leaves only the identity $1$ fixed. If $f \in Aut(G)$ is fixed-point free, then $G=\{x^{-1}...
2 votes

Does $G/K \cong H$ imply that $G \cong H\times K$ for normal $H,K$?

The statement does not hold as written; it requires further assumptions, such as the ones given by reuns. For a counterexample, consider $G=C_4$ the cyclic group of order $4$, and let $H$ and $K$ be ...
0 votes

Prove that $f(H)=\{y\in G∶y=f(x)\text{ for some }x\in H\}\le G.$

Note that all you need is $f$ to be a group homomorphism, as the injectivity and surjectivity of $f$ (as a map with codomain $G$!) don't play any role in proving the claim. In fact, $e_K\in H$ and $f(...
  • 1,295
4 votes
Accepted

Does $G/K \cong H$ imply that $G \cong H\times K$ for normal $H,K$?

If $H,K$ are two normal subgroups of $G$ such that the map $H\to G/K,h\mapsto hK$ is an isomorphism then obviously $H\cap K=\{1\}$ (the map is injective) and $G = \{ hk, h\in H, k\in K\}$ (the map is ...
  • 73.9k
3 votes
Accepted

Is the affine algebraic set defined by the ideal $\langle xy\rangle$ isomorphic to the affine line?

I am assuming that you work in the language of schemes. The line and $V(xy)$ are not isomorphic. Here are some options: You can show that $V(xy)$ has two irreducible components, while $\mathbb A^1$ ...
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1 vote
Accepted

Subgroup of matrices is isomorphic to a given semidirect product

Let the two subgroups be $$G_1=\left\{ \begin{pmatrix} 1&b\\&1\end{pmatrix}\middle| b\in K\right\}\ \text{ and }\ G_2=\left\{ \begin{pmatrix} a&\\&d\end{pmatrix}\middle| a,d\in K^*\...
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6 votes
Accepted

Isomorphism on a Sylow-$p$-subgroup of a group $G$.

That's certainly not true -- take any group with more than one Sylow $p$-subgroup. Sylow's second theorem says all the Sylow $p$-subgroups are conjugate, so in particular there's an inner automorphism ...
  • 27.2k
2 votes

If $G \oplus H$ is isomorphic to a proper subgroup of itself, then must the same be true of one of $G$ and $H$?

This seems to be an open question. Lets start by rewriting it. A group $G$ is called coHopfian if any injective endomorphism $G\to G$ is an isomorphism. Then the question asks: Question. Suppose $A$ ...
  • 28.7k
2 votes
Accepted

Let $\phi:G\to H$ be a surjective hom. of groups. Let $\sigma:H\to G,\phi\sigma=id_H.$ Show $G\cong\ker(\phi)\rtimes H.$

Since $\phi \circ \sigma = \mbox{id}_H$, we can conclude that $\sigma$ is injective. (Recall that for arbitary functions $f: X \to Y$ and $g: Y \to X$, $g \circ f$ being bijective implies that $f$ is ...
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1 vote
Accepted

Is there a group with a proper subgroup such that the quotient is isomorphic to itself?

Take your favourite group $\tilde{G}$ and let $G = \prod_{n=1}^{\infty} \tilde{G}$. Let $f : G \to G$ be the shift $f(x_1, x_2, \ldots) = (x_2, x_3, \ldots)$. Then $f$ is a surjective group ...
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