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5

Since $x+x = 0$, the set $H := \{0,x\}$ is a subgroup of $G$ of order $2$. Lagrange's Theorem implies that $|G| = |G:H||H| = 2 |G:H|$, so $|G|$ is even.

4

Suppose that $\rho : G\longrightarrow {\rm Aut}(X)$ is a group action with $G$ simple. If $\rho$ is nontrivial, the fact that $G$ is simple means $\ker\rho=0$, that is, the action is faithful. If $|X|=n$; then $G$ is a subgroup of ${\rm Aut}(X)$, which has $n!$ elements. Now if $n<5$; then $n!=1,2,6,24$, so that the order of $G$ is a divisor of $24$. But ...

3

All you need is Cauchy's theorem: if $p$ is prime and $p|o(G)$ then there exists an element of order $p$ in your group. $x^2=e$ for all $x$ says everything has order at most 2, so no other prime can be a divisor of the order of $G$ (or we would have an element of that order, which we don't). Fundamental theorem of arithmatic then gets you that the order ...

3

Let $aC(G)$ be a coset which generates $G/C(G)$, so that $$G=\bigsqcup_{n\geq 0}a^nC(G).$$ Consider any two elements $g,h\in G$. We can write them as $g=a^nx$ and $h=a^my$ where $x,y\in C(G)$. Then $gh=a^nxa^my=a^{n+m}xy=hg$, where we have commuted $x$ and $y$ past everything (because they belong to the center). Hence $G$ is abelian.

3

If your group $G$ of order $8$ has no elements of order $4$, then either it has an element of order $8$ (so $G$ is cyclic, in particular abelian) or every nonidentity element of $G$ has order $2$; in the latter case, $(xy)^2 = e = x^2y^2$ for all $x,y\in G$, so $G$ is abelian.

3

Your idea is right on target! Let $C_a = \{a, a+x\}$ for $a\in G$. Prove that when $a$ varies over $G$, the $C_a$ form a partition of $G$: their union is all of $G$ and $C_a$ and $C_b$ are either disjoint or equal. Finally, prove that $C_a$ has exactly two elements. The fact that $G$ is abelian is immaterial. This is the proof of Lagrange's Theorem for ...

3

Yes, you will need to use the fact that $G$ has an element of order $2$. Let $x \in G$ have order $2$, and $y \in G$ have order $n$. Here is a hint. Using the fact that $G$ is abelian and the fact that $\gcd(2, n) = 1$, show that the order of $xy$ is $2n$, whence $G$ must be cyclic. Here is a proof that any group of even order must have an element of order ...

3

No, they're not equivalent. For example, let $S$ be any set, and let $+$ be the operation on $S$ given by $a + b = b$ for all $a,b\in S$. This satisfies (2) but not (1). The issue here that you might be overlooking is that (1) says that there exists a single element $0$ which satisfies $a+0 = a$ for all $a$, while (2) only guarantees that for each $a$ there ...

3

Yes, this is true. Let $S$ be a basis for $A$; then for any group $G$, there is a natural isomorphism $A\otimes G\cong G^{\oplus S}$, where $G^{\oplus S}$ denotes the direct sum of copies of $G$ indexed by $S$, i.e. the group of all functions $S\to A$ which are $0$ at all but finitely many points (this is because $A\cong\mathbb{Z}^{\oplus S}$, ...

2

The point is you're working modulo 7, so you need, for all $x<7$, to find a $y$ such that $xy\equiv1\mod7$. $\mathbb{Z}_7^\ast=\{1,2,3,4,5,6\}$. So here goes: \begin{gather*} 1\cdot1=1; \\ 2\cdot4=8\equiv1\mod7; \\ 3\cdot5=15\equiv1\mod7; \\ 6\cdot6=36\equiv1\mod7 \end{gather*} Summarizing: $$\begin{matrix} \text{Element} & \text{inverse} \\ 1 ... 2 Hint: If a,b are two elements of order 2, how many elements are there in the set \{1,a,b,ab\} ? 2 As frank000 points out, any cyclic group of odd order is 1/2-convex. So if you have a non-cyclic 1/2-convex group and take an element of odd order, the subgroup generated will be a proper 1/2-convex subgroup. For an example, let n be an odd positive integer and consider \mathbb G=Z_n^2. This is a 1/2-convex group, and the subgroup H generated ... 2 A group of order p^2 with p a prime is always commutative : Because the cardinal is a power of p, the centre of G is non trivial : indeed, |Z(G)|\equiv 0 [p], so cannot be 1. Then, if G was not abelian, you would have Z(G)\subset G not to be an equality. Like |Z(G)| divides |G|, it is 1, p, or p^2. It is not 1 because the centre ... 2 Check that f:G\to G given by f(x)=axa^{-1} is an isomorphism where a\in G And in an isomorphism order of an element is preserved 2 It probably means a group action. The symmetric group S_3 (containing all permutations of 3 elements) is noncommutative and acts on 3 elements nontrivially, which means that not every group element performs the identity action. However, it is not simple, so we're not there yet. A further hint: see the list of finite simple groups 2 Here's a completely different proof. As you said, such a group is always abelian. Then G becomes a vector space over the field with two elements \mathbb{F}_2 = \{0,1\}, by setting 0 \cdot x = e and 1 \cdot x = x (verify that this defines a vector space, in particular e = 0 \cdot x = (1+1) \cdot x = x^2 is satisfied). Since G is finite, it is ... 2 Write [x,y] = x^{-1} y^{-1} xy and x^y = y^{-1} xy. Then we have the identities [x, yz] = [x,z] [x,y]^z, [xy,z] = [x,z]^y [y,z] and [x,y]^g = [x^g, y^g]. For example by applying the first two identities, you can see that it is enough to show that the set \{[a,b] : a \in A, b \in B\} is closed under conjugation. Let a \in A and b \in B. Then ... 1 This is a slightly different method. If G is trivial, the statement trivially holds. Let G be a non trivial group and suppose |G| = p_1 ^{n_1} \cdots p_k ^{n_k} be the primary decomposition of |G|. Let H \in Syl_{p_i}(G). Every element of H is either trivial or has order 2 by assumption. Pick a non trivial element x \in H Then |x| =2 ... 1 HINT for the 1st part: You have already proven that o(gag^{-1})\leq o(a), this for all a,g\in G. In order to show the other inequality, notice that$$a=g^{-1}(gag^{-1})gso you can apply the previous case with g 'equal' to g^{-1} and a 'equal' to gag^{-1}. Here you have more details about it: We have proven that for all b,h\in G: ... 1 A slight variation on pre-kidney's answer: Suppose a is a cyclic generator of G/C(G). Then for any g\in G we have g=a^nx where x\in C(G), so ag=aa^nx=a^nax=a^nxa=ga, so a\in C(G) and G=C(G). Hence G is abelian. 1 The one that comes to mind is - \begin{align} &R = \mathbb{Z}, V = \mathbb{Z}, M = \mathbb{Z}/2\mathbb{Z}, D= End_{\mathbb{Z}}(M)\\ &\implies Hom_{R}(M,V) = Hom_{\mathbb{Z}}(\mathbb{Z}/2\mathbb{Z},\mathbb{Z}) = 0\\ &\implies Hom_{R}(M,V) \bigotimes_{D}M = Hom_{\mathbb{Z}}(\mathbb{Z}/2\mathbb{Z},\mathbb{Z}) \bigotimes_{D}\mathbb{Z}/2\mathbb{Z} ... 1 You can do this by a counting argument. Let  x be the element of order 5, and let y\in G\setminus \langle x\rangle. If  y has order  5, then \langle x\rangle \cap\langle y \rangle= \{e\}, since if y^k=x^j for some k, then by raising to the power of k^{-1}\pmod 5, we get y\in \langle x \rangle. But there are only 9 elements in \langle ... 1 Suppose the contrary. Since G is abelian \operatorname{Aut}G contains an element of order 2 (f:G\to G, f(x)=x^{-1}), so 2\mid p^2 and hence p=2. (However, if x^{-1}=x for all x\in G, then G is isomorphic to a direct sum of copies of \mathbb Z/2\mathbb Z and unless |G|=2 it automorphism group is not abelian.) But in this case ... 1 Second question first: the coproduct of morphisms is the same thing as the pushout. You're right that the coproduct of objects is the direct sum. Your discussion of the fiber product is quite confused: it's not actually clear to me what you're trying to show. In a literal sense, it does not require proof that the fiber product exists, at least, if you ... 1 Note that if G/H \cong \text{im }\phi = \langle aH\rangle, then since: |G/H| = p^{n-1}, it follows that a^{p^{n-1}}\in H, but no lesser power is. In particular, the order of a in G is at least p^{n-1}. Could it be that the order of a is p^{n-1}? Well that would mean that (a^{p^{n-2}})^p = e, that is a^{p^{n-2}} \in H, so that: ... 1 We know that for x \in G we must have x^{|G|} = 0, therefore if x has order 2 and |G| was odd, then x^{|G|} = x^{2k+1} = x (for some k). As we assumed x \neq 0 this proves that |G| is even. EDIT: If we do not want to use Lagrange's theorem, we can combine the answer linked in the comments and my answer to get the following: Let G = \{ ... 1 You have to verify the properties that define a group: associativity: we have to check \bigl((a,b)*(a',b')\bigr)*(a'',b'')=(a,b)*\bigl((a,b)*(a',b')\bigr)*(a'',b'') for any a,a',a'',b,b',b''. The left-hand side is equal to \begin{align*}(a+a',bb')*(a'',b'')&=\bigl((a+a')+a'', (bb')b''\bigr)=\bigl(a+(a'+a''), b(b'b'')\bigr)\\ &= (a,b)*(a'+a'', ... 1 G has an element of order n, say g. Also, since |G|=2n, G has an element of order 2 (by Sylow Theorem), say h. Since G is commutative,\langle g\rangle\langle h\rangle =\langle g\rangle \times \langle h\rangle. Furthermore, this group has exactly $2n$ elements, so it is equal to $G$. Therefore $G$ is cyclic.

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