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The main theorem on finite abelian groups is that they can all be written as direct sums of cyclic groups of prime power order (this is called the elementary divisor decomposition). So write $A=\oplus_{i=1}^m \mathbb{Z}/{p_i^{e_i}}$ and $B=\oplus_{j=1}^n \mathbb{Z}/{q_j^{f_j}}$ so that $A \otimes B = \oplus _{1 \leq i \leq m, 1 \leq j \leq n} ... 4 Every f.g. abelian group is the direct sum of finitely many finite cyclic groups and a f.g. torsion-free (hence free) abelian group.$\mathrm{Ext}_\mathbb{Z}^1 (-, \mathbb{Z})$preserves finite direct sums, so it suffices to show that$\mathrm{Ext}_\mathbb{Z}^1 (M, \mathbb{Z})$is a finite group if$M$is a finite cyclic group. In fact, for all positive ... 3 Suppose$A=\Bbb{Z}/m\Bbb{Z}$and$B=\Bbb{Z}/n\Bbb{Z}$. If$\gcd(m,n)=1$then$n$is a unit in$A$and it annihilates$B$. For any pure tensor$a\otimes b\in A\otimes_{\Bbb{Z}}B$we have $$a\otimes b =(nn^{-1}a)\otimes b=(n^{-1}a)\otimes(nb)=(n^{-1}a)\otimes0=0.$$ As all pure tensors are trivial and$A\otimes_{\Bbb{Z}}B$is generated by them, it follows that ... 3 This answer follows M. Bommireddy et al, Arch. Math. 42, 573-576 (1984). Consider$a,b\in Q$. Even in a non-abelian group,$0+b=b+0$, so the only non-trivial case is$a\neq 0$. Since$(Q\setminus \{0\},\cdot)$is a loop, there exists$s\in Q\setminus \{0\}$such that$sa=b+a-b$. Define$x_1$and$x_2: $$1\cdot x_1 = sx_1+b\qquad (1)\\ x_2 := x_1+a ... 3 Short answers Q1: Suppose A,C are given abelian groups. Which groups B can fit inside an exact sequence A\to B \to C? A1: A lot! All this means is that B has a subgroup K=\operatorname{im}(A\to B) so that B/K is isomorphic to a subgroup Q =\operatorname{im}(B\to C) of C. So we are looking for B such that B has a subgroup K ... 3 Since G has odd order, that means G is finite. So \phi can be injective if and only if it is surjective. To formally prove surjectivity, argue by contradiction. Suppose there is a y \in G such that \forall{x} \in G, \phi(x) \neq y. That implies there are two distinct x_{1}, x_{2} such that \phi(x_{1}) = \phi(x_{2}) by injectivity. 3 Here is the case for abelian groups: Theorem: If M is an abelian group with M \otimes M \otimes M = 0, then M is a divisible torsion abelian group and M \otimes M = 0. The proof is pretty standard abelian group theory. Basic subgroups are probably not well known outside of that theory (at least I never learned about them over rings that weren't ... 3 The homomorphism theorems imply that when N is a normal subgroup of G, there is a bijection between the subgroups of G containing N and the subgroups of G/N given by$$ H\mapsto H/N $$In this correspondence, normal subgroups correspond to normal subgroups. When N=G', the quotient G/G' is abelian, so each of its subgroups is normal. So, since ... 2 Yes, it is true for finite groups and there are counter examples in infinite groups. case1: Let G be a finite group; Assume that G is nonabelian. Let g\in Z(G) and g\neq1 then C_g=G then it is true for all element which is a contradiction since we assumed that G is nonabelain. So, We must also assume that Z(G) is a trivial group. Let's ... 2 By the classification of finite abelian groups we know$$ G \cong \mathbb{Z}/(n_1) \oplus \mathbb{Z}/(n_2) \oplus \cdots \oplus \mathbb{Z}/(n_k)$$where n_i divides n_{i+1}. If k\geq 2 i.e. there are at least two summands in this decomposition, then the condition fails: x^{n_1}=e has n_1 solutions from \mathbb{Z}/(n_1) \oplus 0 \oplus \cdots ... 2 So you're basically asking how to prove that B)\impliesC) and C)\impliesD). \bbox[5px,border:2px solid #4B0082]{B)\implies C)} Note that$$\begin{align} \forall x,y\in G\left((xy)^{-1}=x^{-1}y^{-1}\right) &\implies \forall x,y\in G\left(y^{-1}x^{-1}=x^{-1}y^{-1}\right)\\ &\implies \forall x,y\in G\left(xy=yx\right)\\ &\implies \forall ... 2 If you define action of\mathbb Z$on$\mathbb Q$by$q\to nq$for$q\in \mathbb Q$and$n \in \mathbb Z $it is trivial to check that$\mathbb Q$is an$\mathbb Z$module. Now, Let$M$be maximal submodule of$\mathbb Q$then$\mathbb Q/M$is an simple$\mathbb Z$module. You can also say that it is cyclic module since by simplicity the module generated ... 2 I belive this is true: we can assume$n_1 \ge n_2 \ge \cdots \ge n_t$ad$m_1 \ge m_2 \ge \cdots \ge m_t$. Assume that$m_j \le n_j$for$j=1, \cdots, r$but$m_{r+1}>n_{r+1}$. Let$G^k$donote that$\{ g^{p^k}|g \in G\}$, just for the simplicity of symbols. Let$l=m_{r+1}-1$. Since$B \le A$, then$B^l \le A^l$. But$A^l=(Z_{p^{n_1}})^l \times \cdots ...

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Your computation with Goursat's lemma appears to be correct. Here is a complete list of all the subgroups of $C_4\times C_2\times C_3$: Order $24$ $C_4\times C_2\times C_3$ $\langle (1,0,0),(0,1,0),(0,0,1)\rangle$ Order $12$ $C_2\times C_2\times C_3$ $\langle (2,0,0),(0,1,0),(0,0,1)\rangle$ $C_4\times C_3$ $\langle (1,0,0),(0,0,1)\rangle$ ...

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There are also many differences between vector spaces and $\mathbb{Z}$-modules. Every vector space has a basis, but not every $\mathbb{Z}$-module. For example, any finite abelian group is not a free $\mathbb{Z}$-module, and the $\mathbb{Z}$-module $\mathbb{Q}$ is not free. Furthermore a free $\mathbb{Z}$-module may have a linear independent set which cannot ...

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We know any finite abelian group is a sum of $p$ primary abelian groups (it's Sylow subgroups), so you may assume that $A$ is a finite abelian $p$-group. Let $x$ be an element of maximal order. Note in this case $|x|=\exp A$. We can assume $A$ is not cyclic (why?), and that $\exp A>p$, else $A$ is a vector space over $\Bbb F_p$ and nonzero elements ...

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Here is a proof using the structure theorem. By taking a primary decomposition of $G$ we can assume that there is a prime $p$ and exponents $a_i \in \mathbb{N}$ such that $n_i = p^{a_i}$ for each $i$. Let $$G = \langle g_1 \rangle \oplus \ldots \oplus \langle g_k \rangle$$ where $g_i$ has order $p^{a_i}$ for each $i$ and $a_1 \le \ldots \le a_k$. Let ...

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As you have noticed, there is an irritating degree shift that pervades this discussion. Consider the simplicial set $B G$ whose $n$-simplices are $n$-tuples of elements of $G$, with face and degeneracy operators are in the bar construction. (In fact, it is the bar construction induced by the free–forgetful adjunction for $G$-sets.) In other words, $B G$ is ...

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The first thing I would do is to reduce to abelian $p$-groups: If $A = G_1 \times \cdots \times G_r$ where each $G_i$ is a $p_i$-Sylow subgroup of $A$, then every subgroup $B$ of $A$ is a product of subgroups $B_i$ of $G_i$'s. (in fact $B_i$ will be the $p_i$-Sylow subgroup of $B$.) So assume $A$ is an abelian $p$-group, say $A=\prod_{i=1}^k ... 1 The rank of a finitely generated abelian group$A$is the dimension of$A\otimes_{\mathbb{Z}}\mathbb{Q}$as a vector space over$\mathbb{Q}$. Since$\mathbb{Q}$is flat as a$\mathbb{Z}$-module, from the exact sequence $$0\to A_1\to A\to A/A_1\to 0$$ you get the exact sequence $$0\to A_1\otimes_{\mathbb{Z}}\mathbb{Q} \to A\otimes_{\mathbb{Z}}\mathbb{Q} ... 1 Suppose S spans G as \mathbb{Z}-module. G is contained in a free \mathbb{Q}-module V; S spans V over \mathbb{Q}, so \operatorname{rank}(V) \leq |S| . Suppose B is a \mathbb{Z}-basis for G; then B is linearly independent over \mathbb{Z} \Rightarrow B is obviously linearly independent over \mathbb{Q} \Rightarrow |B| \leq ... 1 Short version: the set S of m for which G\{m\} is not finite is always a union of prime ideals, and the set P of m such that G\{m\}=G is an ideal, so if P=\mathbb{Z}\setminus S, then either S=\mathbb{Z} or for some prime p, G=G\{p\} is an infinite elementary abelian group. The ideas of multiplicatively closed sets and prime ideals are ... 1 This isn't an answer, but I can't fit this properly in the comment section. At this point, I have a group of prime power order, in particular C_n \oplus C_{p^k}. My textbook (Hungerford, 2nd edition) is suggesting as a hint to use Theorem 7.8, which says Let G be an additive group and let a \in G. If a has infinite order, then the ... 1 One key step in this problem is the fact that U_{ab} \cong U_a \times U_b, when a and b are co-prime. This comes from the Chinese Remainder Theorem. The order of U_a is \phi(a) and the order of U_b is \phi(b). In most cases, both \phi(a) and \phi(b) are even and so m=lcm(\phi(a),\phi(b))<\phi(a)\phi(b)=\phi(ab) is an exponent for ... 1 \newcommand{\Z}{\mathbb{Z}}Consider the case of finite abelian groups. Let M = \Z/2\Z, N = \Z/n\Z. Then the rank of F(M \times N) is 2n, so any subgroup has rank at most 2n. But already there are 2\binom{n}{2} elements of the form B_{m,n,n'}, which is > 2n if n > 3, in which case the B_{m,n,n'} cannot be linearly independent. 1 Hint: Let G be a finitely generated abelian group. Then G \simeq \mathbb{Z}^n \times \mathbb{Z}_{n_1} \times \cdots \times \mathbb{Z}_{n_r} for some n,n_1,\dots,n_r. Now consider the epimorphism$$G \twoheadrightarrow \mathbb{Z}_k^n \times \mathbb{Z}_{n_1} \times \cdots \times \mathbb{Z}_{n_r}$$with$k$arbitrarily large. 1 Since$G$is not abelian, there exists$g \in G \setminus Z(G)$, and hence there exists$h \in G$with$h^{-1}gh \ne g$. So the inner automorphism$\kappa_h$does not fix$g$. Let$x$be the transposition$(1,g) \in {\rm Sym}(G)$. Then$(1,g)^{-1}\kappa_h(1,g)$does not fix$1$and so cannot be lie in${\rm Aut}(G)$. 1 A sketch of a proof is as follows: First prove that if$H,A_1,\ldots,A_n\leq G$are abelian subgroups and$G=H\cup\bigcup_i A_i$, then if$H$has infinite index, then$G=\bigcup_i A_i$. This is the hard part (though I would appreciate a simpler explanation), I believe: to show this, choose arbitrary$h\in H$and: Notice that there is some$A_i\$ such ...

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There is an unpublished result by Reinhold Baer (see Theorem 4.6 in D.J.S. Robinson, Finiteness conditions and generalized soluble groups, Part I, Springer Verlag, New York, 1972.) Theorem A group is central-by-finite if and only if it is the union of finitely many abelian subgroups.

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