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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 No, take M=\mathbb Z, L=2\mathbb Z. 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.