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0

According to this paper (which I just found with a Google search, so it'd probably be useful to look up the citation it lists), the Schur multiplier of $GL_n(\mathbb{F}_p)$ is trivial for $p\not = 2$. The paper mentions that it was "computed only recently" (the paper it lists was published in 2008), though that might just be the general case rather than $p = ...


1

Your proofs are fine, but for one typo: under the conventions $[a,b] = a^{-1}b^{-1}ab, a^c = c^{-1}ac$, the formulas you implicitly use are: $$[a, bc] = [a,c][a,b]^c, \qquad [ac, b] = [a,b]^c[c,b]$$ Applying the first gives $[x,yy^s] = [x,y^s][x,y]^{y^s}$ (rather than $[x,y^s][x,y]^y$), but since $[x,y] \in Z(G)$, $[x,y]^{y^s} = [x,y]^y = [x,y]$.


0

Here is the list of examples we have which were directed by the comments thus far: $PSL_n(K)$ for when $K$ is an infinite field and $n\geq 2$ 1 The finitary alternating group $A(\kappa)$ for any infinite cardinal $\kappa$ 2 REFERENCES This entry at Groupprops shows that $PSL_n(K)$ is actually simple for all $n\geq 2$ and any $K$ except for $PSL_2(\Bbb ...


3

Note that $b^{n} = a b a^{-1} b^{-1} = (b a b^{-1} a^{-1})^{-1} = a^{-n}$. So $b$ and $a$ commute with the commutator $[a, b] = a b a^{-1} b^{-1}$, which is therefore central, and so $a^{n}$ and $b^{n}$ are also central. It follows that $$ 1 = [a^{n}, b] = [a, b]^{n}, $$ so $[a,b]$ has order at most $n$. Clearly $G' = \langle [a, b] \rangle$. Addendum ...


0

I would check the order the group generated by the commutator $[a,b]$ (and show that it commutes with $a$ and $b$).


1

As Michael stated before, you have included that information in the question itself, if $b_{i}=a_{2}$ for some $i$, then $\tau^{i-1}(a_{1})=b_{i}=a_{2}$ or $\tau^{k}(a_{1})=a_{2}$ for $k=i-1$.


2

It is the canonical group homorphism $G \to G/N$, $g \mapsto gN$ which actually belongs to the definition of the quotient group $G/N$. It is usually called the projection or quotient map. Often the term "reduction map" is used for maps which are derived from the quotient map. For instance, there is a reduction map $\mathrm{GL}_n(\mathbb{Z}) \to ...


2

Cayley proved that every group, even groups with an uncountable number of elements, can be represented by the permutation group of some underlying set. Maybe this Wikipage can help you: http://en.wikipedia.org/wiki/Cayley_group


2

Here is how I see it (without reference to "messy" product representations). If we partially order the lattice of subgroups of $G$ by inclusion, then $H \vee K = \langle H,K\rangle$, the subgroup generated by the set $H \cup K$. By definition this is the smallest subgroup (via our partial order) of $G$ that contains both $H$ and $K$. Now for any function ...


3

That we have $f(X\vee Y)\subseteq f(X)\vee f(Y)$ should be clear (the image of a product of elements in $X$ or $Y$ is a product of elements in $f(X)$ or $f(Y)$). For the other direction, let $g\in f(X)\vee f(Y)$. Then $g=g_1\cdot\cdots\cdot g_n$ where $g_i\in f(X)$ or $g_i\in f(Y)$. Thus each $g_i$ is of the form $f(x_i)$ where $x_i\in X$ or $x_i\in Y$. ...


0

This is great! So here's some feedback on the presentation: Proof of (G1): This reads backwards in my opinion: "let $A=\begin{bmatrix}a&0\\c&d\end{bmatrix}$ and $B=\begin{bmatrix}e&0\\g&h \end{bmatrix}$ where $A, B \in S$..." I suggest writing: "let $A, B \in S$. Then, by definition, $A=\begin{bmatrix}a&0\\c&d\end{bmatrix}$ ...


1

You're on the right track with the first part (but, as suggested in the comments, you should be careful with your justifications). For the second part, I'm not sure your approach will be fruitful: even if you show $HK$ is a subgroup equal to $G$, it's not clear to me that this is useful. To help you out, I'll sketch a solution to a similar problem. You can ...


2

It is enough to prove $X_{0,n}$ is nilpotent for all $n$, so suppose inductively that $X_{0,n}$ is nilpotent for some $n$, with lower central series $X_{0,n} = L_1 > L_2 > \cdots L_{c+1}=1$ for some $c$. For $1 \le i \le c$ and $0 \le j \le n$, let $K_{i,j} = (X_{j,n} \cap L_i)L_{i+1}$. Then, since $[X_{n+1},X_{j,n}] \le X_{j+1,n}$, we have ...


0

In general, if $G$ acts on a set $S$ from the right via $(x,g) \mapsto x \cdot g$, then we obtain an action of $G$ on $S$ from the left via $g \cdot x := x \cdot g^{-1}$. Let $*$ be the group multiplication in $G$. Then $G$ acts on $|G|$ (the underlying set of $G$, don't confuse it with $G$) from the right via $x \cdot g := x * g$. It follows that we obtain ...


4

Let $G$ be the group in question. In $G$, we have: $x^2 = y^2x^2y$ (1), and $yx^{-1}y^2=x^7$ (2) $\Rightarrow y^2=xy^{-1}x^7$ Substitute in (1) $x^2=xy^{-1}x^7x^2y \Rightarrow yxy^{-1} = x^9$ (3) (3) $\Rightarrow yx^{-1} = x^{-9}y$, substitute in (2) to get $x^{-9}y^3 =x^7\Rightarrow y^3 = x^{16}$. Since the element $y^3=x^{16}$ commutes with both ...


1

Take a look in the abstracts of the 16 workshop in Bedlew (non coummutativ harmonic..) 6-12 of 7


2

Even easier than the free group comment. Pick $n\in\mathbb N, n>1$. For each $k\in\mathbb N$, write $k$ in base $n$, yielding an infinite sequence of digits from $0$ to $p-1$. Treat this as an element of $\mathbb Z/p\mathbb Z$. Add two numbers by the same algorithm you'd normally add by, but don't carry the ones. Then you get groups of the form: ...


3

I think the answer is no. Let $G$ be a free group generated by $a, b$. Then $G=\langle a \rangle * \langle b\rangle$. Let $\phi: a\rightarrow b, b\rightarrow a$. Then $\phi$ is an automorphism of $G$. Let $g\in G$ with $|g|=2$, then $g=a_1b_1$ where $a_1\in \langle a \rangle$ and $b_1\in \langle b \rangle$. Now $\phi(g)=b_1a_1\sim g$. But $\phi(a)=b \not ...


2

For any integer $n$, take the direct sum of $\mathbb{Z}_n$ and countably many copies of $\mathbb{Z_2}$. We can find an explicit injection by looking at integers of the form $$2^kp_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}\cdots,$$ ("infinite" product), where $0\le k\le n-1$, the $p_i$ are the odd primes, and $e_i=0$ or $e_i=1$ for all $i$, with only finitely many ...


2

I like the answer from Morgan O, but you could also get this from the following fact, which you may already know (and is not too hard to prove): the derived subgroup $F^{\prime}$ consists of those elements of $F$, thought of as (reduced) words on $X$, for which the exponent sum of each generator from $X$ is equal to zero. Now, if $x,y\in X$ and $xF^{\prime} ...


3

Note that $xF' = yF'$ if and only if $y^{-1}x\in F'$. By the universal property of the commutator subgroup, any morphism $\varphi$ of $F$ into an abelian group factors uniquely through $F/F'$. That is, given $\varphi: F \rightarrow A$ for $A$ abelian, there exists a unique $\bar\varphi: F/F' \rightarrow A$ so that $\bar\varphi \circ \pi =\varphi$, where ...


2

First show it has eight elements by writing down $I, A, A^2, A^3, A^4, ...$ and $B, B^2, \ldots$ and $AB, (AB)^2, \ldots$, etc.; that's what's meant by the group "generated by these matrices: all arbitrarily long products of all possible powers of the matrices. You'll find that every sequence of $A$s and $B$s can be simplified to one of just 8. As an ...


2

I suspect that for the $\Rightarrow$ direction, you are expected to make use of the antecedent: namely, the fact that $\sim$ is an equivalence relation on $G$, meaning, it is reflexive, symmetric and transitive. It is true that for non-empty subset $S$ of $G$ such that $\forall a,b\in S; ab^{-1}\in S$ suffices to affirm that $S\leq G$, but unless you've ...


4

No. For example, let $G$ be the additive group of sequences $(a_n)$ of elements of $\mathbb{Z}/2\mathbb{Z}$ such that $a_n=0$ for all but finitely many $n$. Then for each of the uncountably many non-zero sequences $(b_n)$ of elements of $\mathbb{Z}/2\mathbb{Z}$ (with possibly infinitely many non-zero $b_n$), there is a subgroup $$\left\{(a_n):\sum ...


1

The order of $x$ need not be equal to $m2^{i+1}$. It could be $q2^{i+1}$ where $q$ is a proper divisor of $m$. Examples are easy to make. For instance, work modulo $5$. Then $2$ has order $1\cdot 2^2$, and $2^{35\cdot 2}\equiv -1\pmod{5}$. Remark: We can say that $2^{i+1}$ is the largest power of $2$ that divides the order of $x$. For suppose to the ...


2

Let $R,A,B$ be groups. Take the semidirect products $G_1 = A \rtimes R$ and $G_1 = B \rtimes R$. Then you can amalgamate $G_1$ and $G_2$ along $R$ and the resulting amalgamated product will split as semidirect product $(A\ast B)\rtimes R$. Also, if $G = A \ast B$ is a free product then $G$ splits as a semidirect product: Let $\beta \colon A \to B$ be a ...


2

use same method; set $\bar H=H\times e$ and $\bar K=e\times K$ since $\bar H\bar K=\bar K \bar H$ then $\bar H\bar K=H \times K$ is a group.


6

First: commutativity of finite sums. $\mathbb{R}$ is a field, and as such, addition is commutative. This means that for two elements $a,b\in\mathbb{R}$, $a+b=b+a$. However, it is possible to extend this result. Consider $a_1,\ldots,a_n\in\mathbb{R}$, and a permutation $\pi$ of $\{1,\ldots,n\}$. Then, by induction, one can easily show that $$ \sum_{i=1}^n ...


6

You can't expect a purely group theoretic proof since the result is fundamentally about convergence. Some analytic element is thus necessary. In fact it is not at all a group-theoretic result. Moreover, I don't know which proof you refer to but the one I know (and I presume it's basically the only proof (up to irrelevant detail)) explains perfectly well ...


0

Use the subgroup test. The set $H\times K$ contains $(e,e)$ since $e\in H$ and $e\in K$. If $(x,y),(a,b)\in H\times K$, can you conclude their product is in $H\times K$. Can you show closure under inverses?


3

Just verify the usual conditions: 1) $e_{G\times G}=(e_G,e_G)\in H\times K$ (i.e. $H\times K$ is a non-empty set) 2) If $(x,y), (z,t)\in H\times K$ then $(x,y)(z,t)^{-1}=(xz^{-1},yt^{-1})\in H\times K$ both of which are immediately verified because $H,K$ are groups.


2

Let $xS^k \in S/S^k$ and $yS^k \in S^{k-1}/S^k$ with $x \in S$, $y \in S^{k-1}$. To show that $S^{k-1}/S^k \le Z(S/S^k)$, we have to show that $xS^k$ and $yS^k$ commute with each other or, in other words, that their commutator is the trivial element of $S/S^k$. Now $[xS^k, yS^k] = [x,y]S^k$. But, since $x \in S$ and $y \in S^{k-1}$, $[x,y] \in [S,S^{k-1}] = ...


0

Here's another method without using Euler's totient function or induction. We rely on a lemma: If $H$ and $K$ are normal subgroups of $G$ such that $H\cap K=\{e\}$, then $$\forall x\in H\;\forall y\in K(xy=yx)$$ Let $G$ satisfy the above conditions, and let's denote its order by $k$. Then since $G$ is finite, we can find an element of maximal order. ...


1

$\mathbb{Z}/n\mathbb{Z}$ is the fundamental group of the 3-dimensional Lens space $L(n,1)$. There is a universal covering map $S^3 \mapsto L(n,1)$ and a deck transformation action of the group $\mathbb{Z} / n \mathbb{Z}$ on the space $S^3$ which acts by isometries of the standard metric on $S^3$. This descends to a metric on $L(n,1)$ which is locally ...


2

I have never heard of an "irreducible action of a group on a set." However, in the case of a group $G$ acting on a vector space $V$ as linear isomorphisms, this means that there are no $G$-invariant subspaces other than $\{0\}$ and $V$. A $G$-invariant subspace is a vector subspace $W \subset V$ so that for any $g \in G$, $g(W) \subset W$.


0

Yes. A finite group $G$ is $p$-solvable if every nonabelian composition factor of $G$ has order coprime to $p$. If $G$ is not solvable, then it has a nonabelian simple group as a composition factor. This nonabelian simple group must have order divisible by some odd prime $r$ (finite groups of order $2^k$ are solvable). Then $r$ also divides the order of ...


3

In general, if $G$ is a group then $[G, G]$ is a normal subgroup of $G$. This is because $g^{-1}[h, k]g=[g^{-1}hg, g^{-1}kg]$. Therefore, $[A_n, A_n]\unlhd A_n$. (In fact, $[G, G]$ is characteristic in $G$, which means that $\phi([G, G])$ for all $\phi\in\operatorname{Aut}(G)$. This is proven using the same argument.) For $n\geq 5$, what does this ...


4

With $g=(123)$ and $h=(145)$ we have $ghg^{-1}=(245)$. Therefore $$ ghg^{-1}h^{-1}=(245)h^{-1}=(245)(154)=(124). $$ Conclusion: If $n\ge5$, then every 3-cycle is a commutator of two 3-cycles. Do you see why? Do you see what it implies? You need to handle $n=1,2,3,4$ separately, but they are easier cases. You remember that $G$ is abelian, iff $[G,G]=1$. Also ...


4

In $\mathbb R^4$ pick regular $n$-gons around $0$ with one in the $(x,y,0,0)$ plane, and the other in the $(0,0,z,w)$ plane. So essentially the isometries of one do not affect the others. A "natural" space with $\mathbb Z/n\mathbb Z$ symmetry is a pyramid with base the regular $n$-gon[*]. In general, if there are spaces $U\subseteq \mathbb R^n$ and ...


0

I would like to write this as a comment but I am not permitted to do that. You can formalize this on various ways. One is that, for a given $n \in \mathbb{N}$, a reduced word of lenght $n$ is a concatenation of either of the forms $g_1 h_1 g_2 h_2 ... g_n h_n$ $g_1 h_1 g_2 h_2 ... h_{n-1} g_n$ $h_1 g_1 h_2 g_2 ... h_n g_n$ $h_1 g_1 h_2 g_2 ... g_{n-1} ...


6

A simpler example is provided by ${\rm GL}(n,\mathbb{C})$ for any $n >1.$ We may choose a transversal to the conjugacy classes which consists completely of upper triangular matrices ( as every matrix in ${\rm GL}(n,\mathbb{C})$ is conjugate to an upper triangular matrix). This transversal generates a proper subgroup of $G,$ as the subgroup it generates ...


1

Rats, I misread the original question. You're looking for $\text{Aut}(\Bbb Z/5\Bbb Z)$, which is indeed isomorphic to $\Bbb Z/4\Bbb Z$. Since this group is cyclic, it's enough to find an element of order $4$, and then all choices are given by powers of that generator. Indeed, multiplication by $-1$ gives you inversion, which is an element of ...


3

No. For example, there exist infinite groups with exactly two conjugacy classes. There might be easier counterexamples though.


2

First we manipulate the relations to show that the group $G$ defined by the presentation is the direct product $\langle x\rangle\times\langle y\rangle$, with $x$ of order a divisor of $8$ and $y$ of order a divisor of $3$. Then, to show that $x$ and $y$ have orders $8$ and $3$ exactly, we exhibit a homomorphism from $G$ onto a group in which the images of ...


5

To the first, yes, to the second, no. I'll address the second question first. An isomorphism can be succinctly characterized as a homomorphism $\phi: G \to H$ which is invertible (it is a nice property of homomorphisms and algebraic structure maps in general that the set theoretic inverse of a bijective homorphism is still a homomorphism). If you accept that ...


8

Take $G=\Bbb Z/3\Bbb Z$ and $H=\{1,x,x^2\}$ where $x^3=1$. Then there are two isomorphisms here, you can send $x\mapsto [1]$ or $x\mapsto [2]$, both are isomorphisms. However, you cannot have $G\cong H, H\cong L, G\not\cong L$, being isomorphic is an equivalence relation, so it's transitive. In particular, if you have isomorphisms: $$\begin{cases}\phi : ...


2

$$x^2=y^2x^2y$$ $$x=x^{-1}y^2x^2y=y^{-1}(yx^{-1}y^2)x^2y$$ $$x=y^{-1}x^9y$$ then we can say that; $$y^2x^2y=x^2=y^{-1}x^{18}y$$ $$y^2x^2=y^{-1}x^{18}$$ $$y^3=x^{16}$$ Now, $$(xy^2)^2=yx^2$$ $$xy^2xy^2=yx^2$$ $$xy^2x(y^3)=yx^2y=y^3x^2y^2=x^{18}y^2$$ $$y^2x^{17}=x^{17}y^2$$ $$y^5x=xy^{5}$$ As a last step; $$x^2=y^2x^2y$$ ...


0

Let $G$ be $\mathbb Z/4\mathbb Z$, and $X=\{a,b\}$. Let $G$ act on $X$ by $1\cdot a = b$ and $1\cdot b=a$ (the whole action is determined by that of $1$). Hence $1$ has no fixpoint, although $2$ acts as the identity, i.e. $\langle 1\rangle$ doesn't act freely.


3

You can do induction on the derived length of $G$. Let $n$ be minimal such that there is counterexample $G$ which is solvable of derived length $n$. Then there is an abelian normal subgroup $N$ of $G$ such that the derived length of $G/N$ is less than $n$, and so the result is true in $G/N$. You need to consider the cases when $0,1,2$ or $3$ elements of the ...


1

You have edited your question to only talk about finite groups. So I will edit my hint, but keep the same flavour: Hint: The action of a group $G$ on an object/set $X$ corresponds to a homomorphism $G\rightarrow\operatorname{Aut}(X)$. If this action is free then this map is injective (these are not equivalent conditions though). So, take a group $G$ which ...



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