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The tool for this is the Smith Normal Form, a kind of Gaussian elimination for PID. For the example $G = \langle s,t,u,v \mid s^{4}t^{2}u^{10}v^{6} = s^{8}t^{4}u^{8}v^{10} = s^{6}t^{2}u^{9}v^{8} = e_G \rangle$ written additively, the matrix is $$\left( \begin{array}{cccc} 4 & 2 & 10 & 6 \\ 8 & 4 & 8 & 10 \\ 6 & 2 & 9 ... 7 In general if an abelian group G is of order pq with p and q being different primes then G is the cyclic group of order pq. In this case, take x\in G such that x\ne e. Let \langle x\rangle be the subgroup of G generated by x. If |\langle x\rangle|=35 then G=\langle x\rangle so is cyclic. If not, then |\langle x\rangle| divides ... 6 By a theorem of Cauchy, G has an element x of order 5 and an element y of order 7. Since G is abelian, x and y commute, and hence the order of xy is the l.c.m. of 5 and 7, which is 35. So G=\langle xy \rangle is cyclic. 6 The \phi,\psi commute, and also the following steps are also OK:$$\phi\psi(g_1g_1)=\phi\psi(g_1)\phi\psi(g_2)=g_3g_4.$$Its not clear in your argument why g_3g_4=\phi\psi(g_2)\phi\psi(g_1)? We are allowed to use commutativity of maps \phi,\psi, and we have to conclude commutativity of g_1,g_2. You may proceed in following directions. (1) Consider ... 5 For one, in studying an abelian group you can utilize a lot of your intuition from doing addition. But more to the point about normal subgroups - One should note that in an abelian group G, every subgroup H is normal, so you can always take the quotient G/H. In a non-abelian group, you need special conditions on H. The definition of a normal ... 4 When he states equality, he actually means that the groups are isomorphic. 3 The generators can be written as e^{\pi i/2},e^{2\pi i/5},e^{\pi i}. Then the values these can generate through multiplication are exactly the values of the form$$e^{2\pi i(m/4 + n/5)}$$for integral m and n (the -1 adds nothing because it can itself be generated from i). These in turn are exactly the distinct values$$e^{2\pi i k/20}$$for ... 3 There is a nice chain of small results which proves this which continues down the path that Groups suggests. If Aut(G) is cyclic, then so is any subgroup of it, in particular Inn(G). Inn(G)\cong G/Z(G) where Z(G) is the center. If G/Z(G) is cyclic, the group is abelian. 3 For the first, gh^2g^{-1}=ghg^{-1}ghg^{-1} 3 For the second, (xH)^2=x^2H=H implies that in G/H every element has order at most 2 and so G/H is abelian. 3 Another way of doing this, which avoids talking about irreducible words and such, is using the universal property: Universal Property of free products: For any group K and any group homomorphisms \phi_G:G\to K and \phi_H:H\to K, there exists a unique homomorphism \phi:G*H\to K extending \phi_G and \phi_H. Thus, to prove that nontrivial elements ... 3 If you know there is a unique inverse to every element a\in G then you know there is a unique inverse to the element gh\in G. Then let's try (gh)^{-1}=h^{-1}g^{-1} as a guess candidate! Note that (gh)(h^{-1}g^{-1})=(h^{-1}g^{-1})(gh)=e. As inverse is unique, it follows that h^{-1}g^{-1} is indeed the inverse of gh!! 3 This isn't even well-defined, because the function \ln is not a bijection (or even always well-defined!) so \exp (and hence +) are not always well-defined. When a^2+b^2=1, the first coordinate of \ln(a,b) is 0, so it does not give an element of the set; maybe you are saying that \ln(a,b)=0 for such (a,b) (with 0 being a special element ... 2 I had a much longer answer prepared, but I think that this should suffice. If you want justification of my argument, I’ll expand this answer. You’re really asking whether the compositum of all finite abelian extensions of K is equal to the compositum of all abelian extensions of K, whether finite or not. But all our extensions are algebraic, and every ... 2 In general, I try to explain abstract algebra as the study of common operations. For example, we use the "+" operation in many contexts: we can add integers, real numbers, complex numbers, vectors, polynomials, matrices, continuous functions, hours in the day, and in many other contexts. All of these "+"'s have similar properties, they are commutative and ... 2 It is not easy, in general, to tell whether two groups are isomorphic. It is, however, quite easy to show that any two groups are "homomorphic." Let G, H be groups. Then I will define \theta: G \to H by \theta(g) = e_H for all g \in G, where e_H is the identity element of H. Then for any g, g' \in G, \theta(g g') = e_H = e_H e_H = \theta(g) ... 2 This question was answered by a comment: The automorphism r↦rs^2, s↦rs has order 4, not 2, and the full automorphism group is nonabelian and isomorphic to the dihedral group of order 8. – Derek Holt Dec 21 '14 at 22:39 Also helpful: Endomorphisms are not the same as automorphisms, BTW. You'd have to figure out which of those are invertible. – ... 2 In fact, if |G| = p^2  then G is abelian. Proof: Since G is a p-group, Z(G) \ne \{e\} (by the class equation). So we have |Z(G)| = p or |Z(G)| = p^2. In the first case, |G /Z(G)| = p so it is cyclic and by your lemma, it is abelian. In the second case G = Z(G) and G is abelian. 2 Note that H=\{x^2 : x \in G \} is not necessarily a subgroup (counterexample G=A_4). Better put H=\langle x^2 : x \in G \rangle, so generated by squares. 2 Think about the Lie algebra as invariant vector fields on the group and recall that flowing by t along X corresponds to multiplying by e^{tX}. The condition [X,Y]=0 means that the flows of X,Y commute, so we have e^X e^Y = e^Y e^X. Thus \exp(\mathfrak g) is Abelian, and the rest of your argument shows that the connected component of the ... 2 As been pointed out you confuse the different notations. As for the elements$$0*0=08*1=04*2=08*3=02*4=08*5=04*6=08*7=0$$1 Hint: Is a and b have orders r and s, and \langle a\rangle\cap\langle b\rangle=\{1\}, ab has order \operatorname{lcm}(r,s). 1 Since G is a p-group, all maximal subgroups must be normal. If M is a maximal subgroup, then G/M, being of order p, is cyclic, hence ...... (can you complete this sentence?) 1 I assume that you would like to use (a^{-1})^n = a^{-n} by saying that if n > 0 then$$ a^m a^{-n} = a^m (a^{-1})^n = a^{m-n} $$by leveraging the case for m,n > 0. The problem is that you can't really do this, because in that case you had two powers with the same basis, but usually a \neq a^{-1}. Instead, what you could go this way. First ... 1 You have to consider a few cases to prove this theorem. In that way, I find it a bit annoying. Perhaps there is a better way, but I do not know of one. Edit: There actually is a straightforward way of shortening this; see A.P.'s comment. You have considered one such case (m,n > 0). Here we will prove (most of) the others. We will call case (i) the ... 1 This is always true. If H is a subgroup of G/K, then let p:G\to G/K be the quotient map and consider the inverse image H'=p^{-1}(H). It is easy to see that H' is a subgroup of G containing K and H can naturally be identified with H'/K. In fact, this argument works for nonabelian groups as well. The converse is also true for abelian ... 1 Your proof is mostly correct; however, there are some (important!) details that should be corrected. We need to establish two facts: A_r \cap A_s = \{0\} and A = A_r + A_s. The second part, where you prove that every element of A is a sum of elements from A_r and A_s is clear. You show that$$a = pra + msa, and it is obvious that $pra \in A_s$, ...

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Here's another answer, motivated by a more recent question that asked for a more "concrete" example. Let $R$ be the polynomial ring $\mathbb Z[x]$. One $R$-module is just $R$ itself, and the other is the ideal $(2, x)$ in $R$. (This ideal consists of all integral polynomials with even constant coefficient.) These two $R$-modules are not isomorphic, as ...

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$o(a)$ is equal to $o(\langle a\rangle)=o(H).$ But, $o(ab)$ is not necessarily the order of $\langle a\rangle.\langle b\rangle=HK$. You may track which argument is not valid in your proof. The statement in the title is wrong. It should be If $o(a)=m$, $o(b)=n$ and if $ab=ba$ then $o(ab)$ divides $lcm(o(a),o(b))$.

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If two groups with binary operation $(G, \cdot)$ and $(H, \star)$ are homomorphic then there $\textit{exists}$ a mapping $\theta:G\rightarrow H$. More precisely, $G$ and $H$ are homomorphic if there exists a mapping $\theta$ such that $\theta(a \cdot b) = \theta(a)\star \theta(b)$ for any $a,b \in G$. Often we don't know what this mapping is, but we find ...

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