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## Hot answers tagged abelian-groups

7

For every elements $a\in G$, consider $$\phi_a:G\rightarrow G$$ sending an element $b$ to its conjugate $a^{-1}ba$. You can easily check that all $\phi_a$ are automorphisms of $G$, called inner automorphisms. By assumption, $\phi_a=id$, which means $$\phi_a(b)=b$$ for every $a,b\in G$, so that $ba=ab$. Alternatively, use $$\frac{G}{Z(G)}\cong Inn(G)$$ ...

5

You can find the number of abelian groups of order $256$ by partitioning $8$: that is, find the number of ways you can sum positive integers to equal $8$: If we partition the number 8, we'll see that there are $22$ distinct ways to sum positive integers to equal $8$ , and hence $22$ abelian groups of order $256$. Prior to Edit of question: I'm afraid your ...

5

You can think of $\mathbb{F}_{p^2}$ as a 2-dimensional vector space over $\mathbb{F}_p$. Multiplication by any non-zero element of $\mathbb{F}_{p^2}$ is then an $\mathbb{F}_p$-linear map on $\mathbb{F}_{p^2}$, and so in this way $\mathbb{F}_{p^2}^\times$ embeds into $GL_2(\mathbb{Z}/p\mathbb{Z})$. Since the multiplcative group of a finite field is cyclic, ...

4

The structure theorem for finitely generated abelian groups says that if $G$ is finitely generated and abelian, then it is a product of finitely many groups of the form $\mathbb{Z}$ and $\mathbb{Z}_q = \mathbb{Z}/q\mathbb{Z}$ where $q$ is power of a prime. If $G$ is finite, then surely it is finitely generated and has no factor isomorphic to $\mathbb{Z}$; ...

4

By Bezout's identity there exist integers $u$ and $v$ such that $pu+qv=1$. This means that $$b=b^1=b^{pu}b^{qv}=b^{pu}a^{pv},$$ so in the abelianization we have $(b^ua^v)^p=b$. On the other hand we also have $$a=a^1=a^{pu}a^{qv}=b^{qu}a^{qv},$$ so in the abelianization we have $(b^ua^v)^q=a$. Therefore the abelianization is generated by (the coset of) ...

4

You are correct about $1$ and $3$, but you should add why this is the case. In addition, exactly one of $2, 4$ is also correct. We can decompose $G = \mathbb Z_{10} \times \mathbb Z_{15}$ and express this uniquely, by the Fundamental Theorem of Finitely Generated Abelian Groups, as $$G \cong \mathbb Z_2 \times \mathbb Z_3 \times \mathbb Z_5 \times \mathbb ... 3 Let (A,+) be an abelian group (written additively: thus instead of x^n we write nx...) For n \in \mathbb{Z}^+ we define A[n] = \{x \in A \ | \ nx = 0\}, the n-torsion subgroup, and A[\operatorname{tors}] = \bigcup_{n \in \mathbb{Z}^+} A[n], the torsion subgroup. Why are these relevant? Because for a,b \in A, if na =nb, then n(a-b) = 0 ... 3 2^4:$$G_1=C_2\times C_2\times C_2\times C_2G_2=C_4\times C_4G_3=C_8\times C_2G_4=C_{16}$$3^2:$$G_5=C_3\times C_3G_6=C_9$$7:$$G_7=C_7$$Now try to built the desired abelian groups of that order. Here I assume |G|=2^4\times 3^2\times 7. Some of them are$$G=G_1\times G_5\times G_7\cong C_{14}\times C_6\times C_{6}\times C_2$$... 2 Well, if the group homomorphisms \gamma_n^* are one-to-one, so you're essentially forming a directed union, then this will be infinitely generated, because any finitely many generators would be in one of the groups of your direct system and therefore can't generate the additional elements in the next group of the system. Presumably, you want sufficient ... 2 Jim's answer gives you everything you need, but let me try to draw the bigger picture in other words. What you are looking at is a short exact sequence, i.e. an exact sequence of the form$$0\to A\xrightarrow{f} B\xrightarrow{g} C\to 0$$This means you have three abelian groups and all arrows are group homomorphisms. Exactness means that at any given ... 2 After the observations above, note that in case G is finite (for infinite groups a similar reasoning applies), it must be an elementary abelian 2-group of rank say n. Then Aut(G) \cong GL(n,2). Hence if |Aut(G)| = 1 then this can only be the case if G is trivial, or n = 1 that is G \cong C_2, the cyclic group of order 2. 2 Yes, you're right. Your statement can be generalized to the multiplicative group K^* of the fraction field K of a unique factorization domain R. Can you see how? In fact, if I'm not mistaken it follows from this that for any number field K, the group K^* is the product of a finite cyclic group (the group of roots of unity in K) with a free ... 2 Let \pi, \phi\in S_{\Omega}. Then \pi, \phi are disjoint if \pi moves \omega\in \Omega then \phi doesn't move \omega. For example, (2,3) and (4,5) in S_6 are disjoint. indeed, \{2,3\}\cap\{4,5\}=\emptyset. Theorem: If \pi, \phi\in S_{\Omega} are disjoint then \pi\phi=\phi\pi. 2 If \Gamma<\Bbb Z^k has index n, then n(\Bbb Z^k/\Gamma)=(0). Therefore \Gamma contains the subgroup (n\Bbb Z)^k. This shows that there's a bijection between the wanted subgroups and the subgroups of$$\Bbb Z^k/(n\Bbb Z)^k\simeq(\Bbb Z/n\Bbb Z)^k$$of index n. Then one can simply apply the structure theorem for finite abelian groups. 2 Take A' \to A \to A/A', the first map is the inclusion and the second is the quotient projection. Take then B' \to B \to B/B', the maps as above. These are both exact sequences, which means \operatorname{im}(\text{inclusion}) = \ker(\text{projection}). You should notice that (\phi', \phi, \phi'') is a morphism of exact sequences, which means that ... 2 Recall that the Fundamental Theorem for finitely generated Abelian groups (which applies of course to finite abelian groups) states that every finite abelian group is isomorphic to, and can be decomposed uniquely as, the direct product of cyclic groups of prime order and/or cyclic groups of order equal to the power of a prime. So each of G, H, K can be ... 2 Consider two elements g,h \in G. You know that (gh)^2 = e  and so you get$$gh gh = e.$$Applying h to both sides on the right gives you$$gh g \underbrace{h h}_\text{$h^2 = e$} = e \cdot h = h.$$So now you have$$ghg = h.$$Aimilarly, applying g to both sides on the right gives you that$$gh\underbrace{gg}_\text{$g^2$} = hg,$$and so you ... 2 Here, you will want to use the theorem of structure of finite abelian groups, that is: Every abelian finite group is isomorphic to a group of the form:$$\mathbb{Z}_{m_1}\times \mathbb{Z}_{m_2} \times\mathbb{Z}_{m_3} \times... \times\mathbb{Z}_{m_n}$$In which m_i divides m_{i-1}, and abviously the product of all m_i has to be the order of the group. ... 1 You can use the structure theorem for such questions. The matrix$$ \left(\begin{matrix} -5 & 1 \\ 1 & -5 \end{matrix}\right) $$has Smith normal form$$ \left(\begin{matrix} 1 & 0 \\ 0 & 24 \end{matrix}\right) $$So G/H\cong(\mathbb{Z}/1\mathbb{Z})\times(\mathbb{Z}/24\mathbb{Z})=\mathbb{Z}/24\mathbb{Z}, which is cyclic. 1 Well, you have grabbed the problematic here, imo: since$$(1,0)+H\neq H\neq (0,1)+H\;,\;\;G/H\;\;\text{is cyclic}\;\iff G/H=\langle\;(1,0)+H\;\rangle$$and then we should get that$$\exists\,m\in\Bbb Z\;\;s.t.\;\;(0,1)+H=m\left((1,0)+H\right)=(m,0)+H\iff(m,-1)\in H\iff (m,-1)=(b-5a\,,\,a-5b)\;,\;\;a,b\in\Bbb Z So we need to solve the diophantine ...

1

To say that the rank of an abelian group is "the number of copies of $\mathbf{Z}$" doesn't quite make sense for abelian groups which are not finitely generated. The rank of an arbitrary abelian group $A$ is by definition the $\mathbf{Q}$-dimension of the $\mathbf{Q}$-vector space $A\otimes_\mathbf{Z}\mathbf{Q}$. As egreg points out in the comments, ...

1

There is a bit of theory to go through to explain how the invariant factors of the matrix you wrote down is actually related to $M,$ you will have to go through your notes to see the precise connection. As for the calculation, perhaps recheck your operations, or post them here so we might we able to say where you went wrong. It turns out that for this ...

1

Sure. First consider the following construction, which works for general $H\leqslant G$. Let $x\in G\setminus H$ and $d$ be the smallest possible integer so that $x^d\in H$. Then $x^n\in H \Leftrightarrow d\mid n$. Let $y\in\mathbb{C}$ be a $d^\text{th}$ root of $f(x^d)$, and note that $y\in\mathbb{T}$ since $f$ goes into $\mathbb{T}$. Let $K=\langle ... 1 As mentioned in the other answer and comments, this is true and works even more generally. Suppose that$A$is a divisible abelian group. If$G$is abelian and$H \leq G$, then any homomorphism$f: H \rightarrow A$can be extended to a homomorphism$g: G \rightarrow A$. As seen in the other answer, this can be proven with Zorn's lemma. In fact, we can ... 1 First, if$p=0$and$q \neq 0$then$G_{ab} \simeq \mathbb{Z} \times \mathbb{Z}_q$, if$p \neq 0$and$q=0$then$G_{ab} \simeq \mathbb{Z} \times \mathbb{Z}_p$. Suppose that$p,q \neq 0$. According to von Dyck's theorem, there exists a morphism$\rho : G_{ab} \to \mathbb{Z}$such that$\rho(a)=q$and$\rho(b)=p$; because any nontrivial subgroup of ... 1 Here is a Computer based approach using GAP which shows the result computationally: gap> Z10:=CyclicGroup(IsPermGroup,10);; gap> Z15:=CyclicGroup(IsPermGroup,15);; gap> s:=DirectProduct( Z10, Z15 );; e:=Elements(s);; r1:=Filtered(e,t->Order(t)=2);; Size(r1); 1 ... 1 The subgroups that m.k. denoted by$G(r)$are often called the Omega subgroups and, in a$p$-group$G$,$G(p^i)$is denoted by$\Omega_i(p)$. The `Agemo' subgroups, denoted by an upside-down$\Omega$, which unfortunately I can't figure out how to do here, are defined as${\rm Agemo}^i(G) := \langle g^{p^i} \mid g \in G \rangle$. The invariants of$G\$ are ...

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