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The Cayley's Theorem solves your question: every finite group is isomorphic to a subgroup of the permutations group.

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As Quang Hong said right away, this does not hold. Making it concrete with the following example. The matrices $$A=\left(\begin{array}{rr}1&1\\0&1\end{array}\right)\qquad\text{and}\qquad B=\left(\begin{array}{rr}1&0\\-1&1\end{array}\right)$$ both have order $p$. Their product $$AB=\left(\begin{array}{rr}0&1\\-1&1\end{array}\right) ... 3 Although you've done it already, let me start with an answer to 1. Taking any a \in S and substituting b=e into the first equation we get$$a * a = a * e * a = e $$and therefore every element of S is its own inverse. This proves number 1, that (S,*) is a group. But it proves more: every non-identity element has order 2. Next, taking any a,b \in ... 2 Suppose that there is a generator a+b\sqrt 2. Then it generates 1 and \sqrt 2, that is, there exist m and n such that$$m(a+b\sqrt 2)=1$$and$$n(a+b\sqrt 2)=\sqrt 2$$This implies that am=1 (hence a\neq 0) and an=0; thus, n=0, a contradiction. 2 Hint: How many of order p^2? How many of order q^4? Multiply. 2 You can show it by using the quotient group as Andrea does, but you can also directly show that n_7=1. A 7-group acts on the normal 4-group by conjugation. This action is an automorphism. The automorphism group of 4-groups have order 2 or 6 and hence a 7-seven group must act trivially. This means that the 7-group commutes with the 4-group. You could show ... 2 If K is a normal subgroup of G of order 4 , you may well argue in H=G/K , and reduce to show, via the correspondence theorem, that in a group of order 2⋅3⋅7=42 there must be a normal subgroup of order 7 . To do this, just consider that the number of 7 -Sylow subgroups in H must divide 42/7=6 , and be congruent to 1 modulo 7 2 I'm not certain that I'm interpreting your question correctly, so this may be rubbish. I think you are asking whether H must be abelian if H is the only subgroup of G of order p^2q, where G has order p^2qr with p,q,r primes and p>q>r. In this case, the answer is "no". Take the non-abelian group H of order 75 = 5^23 (so p=5 and ... 2 I'm going to speak more generally of A modules for an F-algebra A, F a field. This would apply to the algebra F[G] for use in representation theory. As Schur's theorem says, D=\mathrm{End}(V_A) is a division ring if V is a simple right A module. Now, for any particular element a\in D and f\in F, fa is indeed another isomorphism of ... 2 Fix x\in H and consider the sequence \{x^n: n\geq1\} since H is stable (closed under multiplication) the sequence is contained in H but H is finite so by the pigeon-hole principle there is n> m so that x^n=x^m. It follows that x^{n-m}=e since n>m implies x^{n-m}\in H. Thus, every element in H has its inverse in H. 2 The idea is to show that elements of P_7 and P_{13} commute. The proof seems to be using that |\mathrm{Aut}(P_7)| = 48 but this is wrong. Because |P_7| = 49 we know that P_7 isomorphic to one of \mathbb{Z}_{49} or \mathbb{Z}_{7} \times \mathbb{Z}_{7} and so |\mathrm{Aut}(P_7)| is either 42 or 48 \cdot 42. The idea of the proof can still ... 2 HINT: If g'\in gH, what can you say about g'H(g')^{-1}? 2 Suppose that |H|=p^\beta, where 1\le\beta\le\alpha. Every subgroup of H is a subgroup of G, so by the induction hypothesis G has subgroups of every order p^\gamma with 0\le\gamma\le\beta. Now suppose that \beta<\gamma\le\alpha+1, and let \delta=\gamma-\beta. |G/H|=p^{(\alpha+1)-\beta}, and \delta\le(\alpha+1)-\beta\le\alpha, so by ... 1 To generalize Jihad's answer: Let G be any group of order p^n. Then G has a subgroup of order p^k for every k=0,\dots,n. Proof by induction: Case 1: G is abelian. This is Jihad's case. Case 2: G is not abelian. Nonetheless (by a standard argument based on the class equation) it has nontrivial center. Say |Z(G)|=p^m with 0<m<n. Then ... 1 Continuing your proof: "...since \;\phi\; is onto there exists \;x\in G\;\;s.t.\;\;\phi(x)=g\; , and thus$$\phi(a)g=\phi(a)\phi(x)=\phi(ax)=\phi(xa)=\phi(x)\phi(a)=g\phi(a)$$1 If I'm not mistaken in understanding of your question. Let's G be an abelian group of order p^n. Does a subgroup of order p^k exist? Yes, it exists: If G is cyclic then subgroup of order p^k is generated by element g = p^{n-k}. It is an implication of the fact that g has order p^k. Every abelian group of order p^n can be represented as a ... 1 Hint By the Chinese Remainder Theorem Z_{1001} \sim Z_{7} \times Z_{11} \times Z_{13}. Therefore U(Z_{1001})\sim U(Z_{7}) \times U(Z_{11}) \times U(Z_{13}). The rest is easy, especially if you know that the multiplicative group of a field is cyclic. 1 Suppose P \le G with |P|=pq^2, |G|=pq^3. As you say, if P has a normal Sylow p-subgroup, then G has subgroups of all possible orders. So suppose that P has a normal Sylow q-subgroup H. Since H is normal in both in P and in a Sylow q-subgroup of G, we must have H \lhd G. Let R be a Sylow p-subgroup of P. If |N_P(R)|>p, ... 1 Let N be a maximal normal subgroup. Then G/N is simple and abelian (because it is smaller than A_5), hence cyclic. N is strictly smaller than G, hence we may assume by induction that itis solvable, hence G is solvable. 1 You never defined what g is, although it is fairly obvious what it is. If g(x)=xa^{-1} then f(g(x))=g(f(x))=x for all x\in G. If a function has both a left and right inverse (or just an inverse) then it is bijective. For a direct proof without defining an inverse, Injectivity If f(x)=f(y) then xa=ya. Multiplying both sides on the right by ... 1 There are 6 cosets Hg_1,\cdots,Hg_6 of H in G, and multiplying each of these on the right by some g \in G will permute these cosets. Thus we have the homomorphism \phi:G\to \text{Sym}(\{gH\}) sending g to its permutation on the cosets. Note that although e does not permute the cosets, all other elements of h \in H will do permute the cosets ... 1 Suppose G has n elements \{g_1,g_2,\ldots,g_n\}... Consider the action \eta : G\times \{g_1,g_2,\ldots,g_n\}\rightarrow \{g_1,g_2,\ldots,g_n\} By this we mean, given g\in G we have \eta _g : \{g_1,g_2,\ldots,g_n\}\rightarrow \{g_1,g_2,\ldots,g_n\} with g\cdot g_i\mapsto gg_i This \eta_g is a permutation... This \eta_g is a permutation ... 1 H=S_n. It's well known fact. Put n=3 for canceling other choices. Observe that, for :$$(1 2 \dots n)^i (1 2) (1 2 \dots n)^{-i} = (i+1 \; i+2)$$Thus, all transpositions of adjacent elements are in the subgroup generated by these two permutations. Since the set of all transpositions generate S_n we see that these two permutations generate the whole ... 1 Proposition Let P be a p-subgroup of a group G, and S \in Syl_p(G), then$$ S \subseteq N_G(P) \text{ iff } P \unlhd S.$$Proof Assume S \subseteq N_G(P). Observe that apparently S \in Syl_p(N_G(P)). Of course, P is a p-subgroup of N_G(P) and must lie in some Sylow p-subgroup of N_G(P), that is P \subseteq S^x, for some x \in ... 1 1- Notice that |H| divides |G| so \gcd(|H|,|G|)=|H|. 2- There's not a group H with three element such that a^2=e since in this case o(a)=2 but 2 doesn't divide |H|=3. 1 Hint: if N_G(S)=S then S has |G:N_G(S)|=9 different conjugates and since M is normal, all of them must lies in M. Can it be possible ? 1 So you found a primitive root in \Bbb Z_7, i.e. which generates its multiplicative group. Therefore, the multiplicative group \Bbb Z_7^* is cylcic, hence \cong\Bbb Z_6. Finally, the isomorphism \Bbb Z_2\oplus\Bbb Z_3\ \to\ \Bbb Z_6 can be given by$$(a,b)\mapsto 3a+2b\,.

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Hint: If $x$ is not in the center, then what contradiction would you get if $|G:C(x)|=1$. Note: the values $|G:C(x)|$ can take are $1,p,p^2,...p^n$

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