# Tag Info

4

Yes, consider $G=H\times S_3$, which is a nonabelian group. We have $$Z(G)=Z(H)\times Z(S_3)=Z(H)\times 1 \cong Z(H)=H,$$ since $Z(A\times B)=Z(A)\times Z(B)$ in general, and $S_3$ has trivial center.

3

You can construct as many counterexamples as you like by taking direct sums of $\mathbb Z$ modulo powers of a given prime. Concretely, a counterexample would be provided by something like $H=\mathbb Z/2\mathbb Z \oplus \mathbb Z/2 \mathbb Z$. By the uniqueness part of the fundamental theorem, we know that $H$ cannot be isomorphic to $\mathbb Z/ 4 \mathbb Z$. ...

3

In fact it is not true in either direction. The infinite dihedral group is abelian-by-finite abelian, but not (finite abelian)-by-abelian. A central product of infinite many copies of $D_8$ is (finite abelian)-by-abelian, but not abelian-by-(finite abelian).

3

Another viewpoint using a bit simpler terms. Containing a copy of $\mathbb Z$ means that it has an element of infinite order, so we have $o(g+H)=\infty$ for some $g\in G$. Now, under the natural projection map, $o(\phi (g))|o(g)$so $\infty |o(g)$, hence $o(g)=\infty$

3

Such groups are exactly the groups that are a (possibly infinite) direct sum of cyclic groups of prime order. Here is a sketch of the proof. First, given a direct sum of cyclic groups of prime order, any subgroup can be canonically decomposed into its $p$-torsion for different primes $p$, and now you use the fact that any subspace of an ...

2

Since $N$ is finitely generated, we have an exact sequence $$M\stackrel{f}{\rightarrow} N\rightarrow \mathbb{Z}^k\times T\rightarrow 1$$ where $T$ is finite and $k \ge 0$. If $f$ is not surjective then either $k\ge 1$ or $T$ is nontrivial. Since tensoring is right-exact, tensoring this sequence with $\mathbb{Z}/p$ gives: ...

2

Let $G$ be an abelian group with subgroups $H, K$ of orders $m, n$ respectively. Then $HK$ is a finite abelian group with $H$ and $K$ as subgroups, so its order is divisible by both $m$ and $n$, hence it is a mulitple of $(m, n)$. The result will now follow from the following fact: If $G$ is a finite abelian group of order $k$ and $l | k$, then $G$ has a ...

2

You're wrong: $\mathbf Q$ is not a free $\mathbf Z$-module, hence it has no rank. To see this tae any two fractions, $\dfrac ab$ and $\dfrac cd$. A linear relation between them is: $$bc\cdot\frac ab-ad\cdot\frac cd=0.$$ More generally, no ring of fractions $S^{-1}A$ can be free over $A$, unless $A=S^{-1}A$.

1

Let $h\in H$ since $h+h+...+h\in H$ this means that $h\mathbb{Z}\subset H$. Now assume there are at most $k\leq n$ linearly independant (over the reals) $h_i\in H$. This means that $H$ contains the lattice generated by them which is a discrete abelien subgroup of $\mathbb{R}^n$ and has rank $k$. Since we assumed $H$ has only $k$ linearly independat vectors ...

1

I assume you mean that a sequence $$0\to G_1 \to G_2\to G_3\to 0\tag{*}$$is short exact. Set $$G_3=\Bbb Z_{a_1}\times \Bbb Z_{a_2}\times \cdots \Bbb Z_{a_n}\times \Bbb Z^m$$ If $G_1=\Bbb Z^n, G_2=\Bbb Z^{n+m}$ and $f:G_1\to G_2$ is given by first multiplying each component with the corresponding $a_i$, then adding $m$ zeroes to the end, the sequence $(*)$ is ...

1

Since $H$ is cyclic hence abelian, we have $H\subseteq C_G(H) \subseteq G$. By order considerations, $C_G(H)$ must be either $H$ or $G$. Suppose $C_G(H) = G$. Then in particular $x$ commutes with every element of $G$, so that $x\in Z(G)$, contradiction.

1

Let $P$ be a Euler-Poincaré map. We have the so called "fundamental" short exact sequences $$0\longrightarrow B_n\longrightarrow Z_n \longrightarrow H_n \longrightarrow 0\\ 0 \longrightarrow Z_n \longrightarrow C_n \longrightarrow B_{n+1} \longrightarrow 0$$ so that $P(C_n)=P(Z_n)+P(B_{n+1})$ and $P(Z_n)=P(B_n)+P(H_n)$ Then \begin{align}\chi(C)&= ...

1

Let $R$ be a commutative unitary ring, $M$ an $R$-module, $N$ a finitely generated $R$-module, and $f:M\to N$ a homomorphism. Suppose that $\bar f:M/\mathfrak mM\to N/\mathfrak mN$ is surjective for every maximal ideal $\mathfrak m$ of $R$. Then $f$ is surjective. $f$ surjective $\Leftrightarrow$ $f_{\mathfrak m}:M_{\mathfrak m}\to N_{\mathfrak m}$ ...

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