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4

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


4

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 $$(a,b)\mapsto 3a+2b\,.$$


1

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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