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9

Yes. The group will be finite. It was given that all the elements of the group $G$ satisfy the equation $M^2=I_n=1_G$. A standard exercise is to show that this implies that $G$ is abelian. All the elements of $G$ are diagonalizable (finite order, eigenvalues $\pm1$ only), so another standard exercise shows that they are simultaneously diagonalizable. So ...


6

If $C$ is a concrete category i.e. a category equipped with a faithful functor $U : C \to \mathsf{Set}$, then the free $C$-object over a set $X$ is by definition an object $F(X)$ which satisfies the universal property $\hom(F(X),A) \cong \hom(X,U(A))$, naturally in $A \in C$. If $C=\mathsf{Set}$ and we choose (what else?) $U=\mathrm{id}$, then clearly we ...


5

Since $G$ has $p+1$ Sylow $p$-subgroups, we have $P = N_G(P)$ for $P \in {\rm Syl}_p(G)$. So if $h \not\in P$, then $h \not\in N_G(P)$, so no nontrivial element of $P$ can centralize $h$. Hence $h$ has $p$ distinct conjugates under the action of $P$. So the $p$ elements that do not have order $p$ must all be conjugate, and so they all have the same order ...


3

A more precise statement of Cayley's theorem states that if $|G| = n$, then $G$ is a subgroup of $S_n$. In this case, $|\mathbb{Z}_6| = 6$, so $\mathbb{Z}_6 \leq S_6$. In particular, it will be the subgroup generated by the $6$-cycle $\sigma = (1,2,3,4,5,6)$.


2

The group $S_n$ has order $n!$ (in group theory it's customary calling order the number of elements in the group). If your interpretation were correct, there wouldn't exist groups with order $3$, $4$, $5$ and so on. But for every natural number $m>0$, there is a group of order $m$, for instance the cyclic group $\mathbb{Z}/m\mathbb{Z}$. Moreover, $S_n$ ...


2

The question is a bit sloppy since you don't specify whether finite index subgroups are meant to be closed. In general this makes a big difference. Anyway here it doesn't: Proposition. Let $G$ be a compact group. Equivalent statements: $G$ has a proper closed finite index subgroup $G$ has a proper finite index subgroup $G$ is not connected To check this ...


2

It can be a subgroup of a group of permutations, not the whole group! For example, $$C_6\cong \langle\; (1,2,3,4,5,6)\;\rangle\le S_6\;$$


2

You took $lcm (n,k)=m$ Now, let $x=a^t \in \langle a^n \rangle \bigcap \langle a^k \rangle$ Then, $n$ divides $t$ and $k$ divides $t$. So, $m=lcm(n,k)$ divides $t$ i.e. $x=a^t\in \langle a^m \rangle$ For opposite containment Let $x=a^t\in \langle a^m \rangle$ Then, $m$ divides $t$ Now, since $n$ and $k$ divide $m$ you have the containment. So ...


2

In $\;S_3\;$ , take $\;a=(12)\;,\;\;b=(123)\;$ . We don't even need to check what the product is to know it can't have order six.


1

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 ...


1

This is an abelian group freely generated by two elements $a,b$ which satisfy $4a-6b=0$. Consider $a'=a-b$ and $b'=2a-3b$. Then $2b'=0$. We have $2a'-b'=b$ and $3a'-b'=a$. Thus we have found new generators. The old relation $4a=6b$ becomes $4(3a'-b')=6(2a'-b')$, which is equivalent to $2b'=0$. Thus, we have the abelian group freely generated by two elements ...


1

Hint: the relation between rotations and reflections looks like this: $$ (f_k \circ c) (z) = f_k(\bar{z}) = \bar{z}e^{2\pi i k/n} = c(ze^{2 \pi i (n-k)/n}) = (c \circ f_{n-k}) (z) $$ Now you should calculate each of the following expressions (for $0 \le k < n$) and use the relation to express the result in the form $f_j$ or $f_j \circ c$ for some $j$ ...


1

Using your first multiplication $\mathrm{AGL}(V)$ acts on $V$ from the left via: $(v,A)*z:=Az+v$ This is an action since $$(w,B)*((v,A)*z)=(w,B)*(Az+v)=BAz+Bv+w=(Bv+w,BA)*z=((w,B)(v,A))*z$$ and feels fairly natural to me.


1

Let $g$ be in the intersection, i.e. $g=(a^n)^r=a^{nr}$ and $g=(a^k)^s=a^{ks}$. This means $a^{nr}=a^{ks}$. Can you take over from here?


1

Take $G = \mathbb{Z}_p$ with $p$ prime, $H = 0$, and $w(x) = x^p$.


1

Two non-trivial examples: Take $G=A_5\times C_2$, $H=A_5$ and take the verbal subgroup to be the commutator subgroup, $w(x, y)=[x, y]$. Then $[A_5, A_5]=A_5$ while $[G, G]=A_5$. Take $S=S_5$, $H=A_5$ and take the verbal subgroup to be the subgroup generated by all squares of elements, $w(x)=x^2$. Then as $A_5$ has index $2$ in $S_5$, $w(G)\leq A_5$. On the ...


1

Just remember that quotients of radicable groups are radicable too and a finite group cannot be radicable. Thus, if $N \leq G$ is such that $[G:N]<\infty$ then $G/N$ is finite, yielding a condradiction.


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

Any group with at most one non-trivial proper subgroup is cyclic: any element not in the non-trivial proper subgroup (or not the identity if there is none) must generate the whole group. Neither of the subgroups $H,N$ of the question can have more than one non-trivial proper subgroup (as with the subgroup itself that would give more than two non-trivial ...



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