Hot answers tagged

17

No. Take $G=S_3$. The abelianization of $G$ is $C_2$. But $C_3$ is a subgroup of $S_3$ and certainly does not inject into $C_2$. The correct universal property of the abelianization is as follows: Let $G$ be a group, let $H$ be an abelian group and let $\phi\colon G\to H$ be a homomorphism. Then there is a unique homomorphism $\hat{\phi}\colon ...


11

In fact, this happens all the time for a nonabelian group $G$. If $G$ is not abelian, then its commutator subgroup $[G,G]$ is nontrivial. Take any $x\in[G,G]$ different from $1$, and let $H$ be the subgroup generated by $x$. Then $H$ is abelian but maps to $0$ in the quotient $G/[G,G]$!


6

This is actually not true. A group for which all subgroups are normal is called a Dedekind group, and non-abelian ones are called "Hamiltonian". The smallest example is the quaternion group $Q_8$. See this MO discussion for more info.


5

$\prod_{i=1}^\infty \mathbb{Z}$ is a counterexample. It is torsion free, not free (Why isn't an infinite direct product of copies of $\Bbb Z$ a free module?), and every nonzero element is only divisible by finitely many integers, so your extra hypotheses hold.


5

The automorphism group, as an additive group, of the $p$-adic integers $\mathbb{Z}_p$ is the same as its group of units, which is isomorphic to $\mathbb{Z}_p\times\mathbb{Z}/(p-1)\mathbb{Z}$ for odd primes $p$. So $\mathbb{Z}_3\times\mathbb{Z}/2\mathbb{Z}$ is isomorphic to its own automorphism group.


4

By the first isomorphism theorem, $Im(f)\cong G/Ker(f)$. Now $N<Ker(f)$ since $N$ is abelian, so $|G/Ker(f)|=|Im(f)|$ is relatively prime to $|Aut(N)|$. But $Im(f)<Aut(N)$, so $|Im(f)|$ divides $|Aut(N)|$. Therefore the only possibility is that $Im(f)=\{e\}$, so that $Ker(f)=G$, i.e. $N$ is in the center.


3

If $G$ and $H$ are groups of coprime order, then every subgroup of $G\times H$ is the cartesian product of a subgroup of $G$ by a subgroup of $H$. (Namely, whenever the subgroup contains an element $(g,h)$, then $g$ and $h$ will have coprime orders, so by the Chinese remainder theorem, the powers of $(g,h)$ will include $(g,1)$ and $(1,h)$ too). Therefore ...


3

Hint: consider $Q_{8}$, the quaternion group.


3

The condition is equivalent to being a torsion $\mathbb{Z}$-module. If $X$ is a torsion module, then any finitely generated submodule is a finitely generated torsion $\mathbb{Z}$-module : by the structure theorem for finitely generated modules over PID, it's finite. If $X$ is not a torsion module, then any non-torsion element generates an infinite ...


3

Let $x\in G$, $x\ne0$; then $\langle x\rangle$ is a (non trivial) cyclic subgroup of $G$. Since $\langle x\rangle\cong G$, we have that $G$ is cyclic.


2

There are (infinitely generated) noncyclic torsion-free groups $G$ such that $Aut(G)\cong {\mathbb Z}/2$, see introduction to J. T. Hallett, K.A. Hirsch, Torsion-free groups having finite automorphism groups. I. J. Algebra 2 (1965) 287–298. and references given there (various examples are due to de Groot, Hulanicki, Fuchs, Sasiada). The paper itself ...


2

Here is a proof that does not use the axiom of choice. Let $e_n\in A$ denote the sequence whose $n$th term is $1$ and all other terms are $0$; we take it as known that any homomorphism $A\to\mathbb{Z}$ which vanishes on each $e_n$ is $0$ everywhere (the standard proofs of this certainly do not use choice). Suppose $A$ were projective. Let $F$ be the free ...


2

Saying that the order of $g$ is $p^k$ for some $k$ is the same as saying that $g^{p^m}=1$ for some $m$. One direction is clear, the other one follows from the fact that if $r>0$ and $g^r=1$, then the order of $g$ is a divisor of $r$. Now, suppose $g$ and $h$ belong to the primary component; then $g^{p^m}=1$ and $h^{p^n}=1$ for some $m$ and $n$. Then ...


2

The given condition implies that $ \phi : x \to x^3 $ is an endomorphism of $ G $. On the other hand, $ \phi(x) = x^3 = e $ implies that $ x = e $ as the group cannot have an element of order 3, so $ \phi $ has trivial kernel, and is therefore an automorphism. $ \mu(x) = g x g^{-1} $ is also an automorphism for any $ g $, which means that $$ g x^3 g^{-1} = ...


2

$ababab = (ab)^3 = a^3b^3 = aaabbb$ $\implies a^{-1}abababb^{-1} = a^{-1}aaabbbb^{-1}$ $\implies baba = aabb$ Use now the fact that the order of the group is not divisible by 3.


2

Hint: $(ab)^3=a^3b^3$ $\Rightarrow (ab)(ab)(ab)=aaabbb$ $\Rightarrow a(ba)(ba)b=a(aa)(bb)b$


2

Just for good measure, let's prove it: $\tanh(x+y)=\tanh(x) * \tanh(y)$ It is well-known that by the $\tanh$ addition identity, $\tanh(x+y)=\frac{\tanh(x)+\tanh(y)}{1+\tanh(x)\tanh(y)}$, which is equivalent to the statement above. $\tanh$ is bijective from $\Bbb{R}$ to $(-1, 1)$ We have $\tanh x=\frac{e^{2x}-1}{e^{2x}+1}$. We have the following: ...


2

Let $G^{\mathrm{ab}}$ be the abelianization and suppose that it contains an element of order $2$. It follows that the map $x\in G^{\mathrm{ab}}\mapsto 2x\in G^{\mathrm{ab}}$ has a non trivial kernel, so —since $G^{\mathrm{ab}}$ is finite— it is not surjective. The subgroup $2G^{\mathrm{ab}}$ of $G^{\mathrm{ab}}$ is therefore a proper subgroup. Now ...


1

The group $\mathbb{Z}^\mathbb{N}$ satisfies your hypotheses but is not free. For a much smaller counterexample, let $\hat{\mathbb{Z}}$ be the profinite completion of $\mathbb{Z}$ and let $\alpha\in\hat{\mathbb{Z}}$ be an element such that $a\alpha+b$ is divisible by only finitely many integers whenever $a,b\in\mathbb{Z}$ with $a\neq 0$ (such an $\alpha$ can ...


1

Suppose $(n,m)= d \not= 1$. Let $(a,b)\in Z/m \oplus Z/n$. Can you show that that $(a,b)^{\frac{mn}{d}}=e$?


1

It's because $G$ is abelian so $(gh)^e = g^e h^e.$ This isn't true for example when $G = S_3$ and you look at the elements of order $1$ and $2$.


1

Your solution does not work. First, $m$ is always different from $0$ because $H(\mathbb Q)$ contains the cyclic subgroup $\{(1,0),(-1,0)\}$. Second, what does it mean that "$\mathbb Q(\sqrt{A})^*$ has no linearly independent elements"? Because actually $\mathbb Q(\sqrt{A})^*$ contains a lot of linearly independent elements! To prove the claim, one can ...


1

I suppose $M=\mathbb Z\times m\mathbb Z$ with $m\ne0$. Then $(\mathbb Z\times\mathbb Q)/M=(\mathbb Z\times\mathbb Q)/(\mathbb Z\times m\mathbb Z)\simeq\mathbb Q/m\mathbb Z$. Write $m=\frac ab$. Now pick an element $x\in\mathbb Q$, $x=\frac cd$. Then $(da)x=cbm\in m\mathbb Z$, so the order of $x$ is finite.


1

No. Set $m=\dfrac ab$. An element $(x, \dfrac cd)$ has finite order in $(\mathbf Z\times\mathbf Q)/M$ if there exists $n\in\mathbf N, z\in \mathbf Z$ such that $$n\frac cd=\frac ab z\iff nbc=adz$$ Just take $n=\dfrac{ad}{\gcd(ad,bc)}$.


1

A very group theoretic way to prove your claim is outlined in the following: Prove that for $n$ and $m$ co-prime, that is $\gcd(n,m)=1$, you have an isomorphism $\mathbb Z_{nm} \cong \mathbb Z_{n}\times \mathbb Z_{m}$ (where $\times$ denote the direct product of groups). Prove that the generators of $\mathbb Z_n \times \mathbb Z_m$ are exactly ordered ...


1

The inverse is given by taking the dual representation, which has character the complex conjugate of the original character. In the $1$-dimensional case this is the inverse of the original character.



Only top voted, non community-wiki answers of a minimum length are eligible