New answers tagged group-isomorphism
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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 ...
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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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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 ...
- 7,248
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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 ...
- 13.7k
4
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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
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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
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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
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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
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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
$$
A(\eta) = \begin{bmatrix}\...
- 20k
0
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Accepted
How to prove that $(G, \cdot)$ structure is an Abelian group where $G$ is a set of matrices determined by a certain rule?
One easily checks that
$$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 ...
- 14.2k
2
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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
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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
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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 ...
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-1
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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}...
- 45.3k
2
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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 ...
- 371k
0
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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(...
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4
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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 ...
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3
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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
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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
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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
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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
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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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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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