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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 (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 $f(g^... • 481 1 vote ### Suppose we have a homomorphism$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 ... • 13.7k 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 $$A(\eta) = \begin{bmatrix}\... • 20k 0 votes Accepted ### How to prove that (G, \cdot) structure is an Abelian group where G is a set of matrices determined by a certain rule? One easily checks that$$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 ... • 14.2k 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 ... • 3,121 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}... • 45.3k 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 ... • 371k 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 ... • 2,721 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^*\... • 644 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 ... • 815 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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