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5

Here is a very elementary proof, which uses Lagrange's Theorem, but not Sylow's Theorem. It is enough to prove that any two subgroups $P,Q$ of order $p$ are equal, because then we must have $gPg^{-1}=P$ for all $g \in G$, so $P$ is normal. So suppose that $P \ne Q$. Then $P \cap Q = \{1 \}$ by Lagrange. They are both cyclic so $P = \{x^i : 0 \le i < p ...


4

We can take either $g$ or $h$ to be identity too , so if $z$ is a 2-element or square of an odd order element then it is obvious. Now , we assume that $n=|z|=2^k.s=t.s$ where $s$ is odd and $t=2^k$ for some $k\in \Bbb{Z^+}$. Now as $(s,t)=1$ there exist $a,b \in \Bbb{Z}$ such that $at+sb=1$ which implies $z^1=z^{at+sb}=z^{at}.z^{sb}$ where $z^{(at)s}=1$ ...


4

$\newcommand{\Span}[1]{\langle #1 \rangle}$$\newcommand{\Order}[1]{\lvert #1 \rvert}$There is no such example. Suppose $$G = \Span{x_{1}} \times \dots \times \Span{x_{n}},$$ where, setting $\Order{x_{i}} = 2^{e_{i}}$, we have $e_{1} \ge e_{2} \ge \dots \ge e_{n} > 0$. Setting $y_{i} = x_{i}^{2^{e_{i} - 1}}$, we have that the involutions of $G$ are the ...


3

Hint: see if you can show that one of these groups is cyclic, whereas the other is not.


3

Take $G=C_2 \times \cdots \times C_2$, with $n \gt 1$ factors. Then $g^2=1$ for all $g \in G$, but $|G|=2^n$.


3

If $g\in C_G(P)$, then the subgroup $Q$ generated by $P$ and $g$ is a $2$-group (every element of $Q$ can be written in the form $g^n p$ for some $n\in\mathbb{Z}$ and $p\in P$ since $g$ commutes with $P$, and $(g^np)^{2^m}=1$ for $m$ sufficiently large since $g$ and $p$ commute and both have order a power of $2$). Since $P$ is $2$-Sylow and $Q\supseteq P$, ...


3

In gap there is a MinimalGeneratingSet(G) command which pulls minimal generating set, in this case it says it is a three element set. There is some more information on this page. The documentation says it is really only efficient for nilpotent, solvable groups, and some special cases (if you have all the correct packages). In this particular case you could ...


3

Claim Let $g_1H, \ldots, g_mH$ be all the cosets of $H$ in $G$. Then any two of the following $g_1Hg_1^{-1}, \ldots, g_kHg_m^{-1}$ intersect only in the identity. Proof. This uses the hypothesis that $H\cap gHg^{-1}=\{e\}$ if $g\notin H$. Suppose $a\in g_1Hg_1^{-1}\cap g_2Hg_2^{-1}$. Then we have $$g_1^{-1}ag_1\in H\cap\ (g_1^{-1}g_2)H(g_1^{-1}g_2)^{-1}$$ ...


3

Hint: Use the fact that $$|HK|=\frac{|H||K|}{|H\cap K|}.$$


2

A group $G$ of order $10$ has a subgroup $H$ of order $5$ (which must be cyclic). As it is of index $2$, $H$ is normal in $G$. If $g\in G$ is of order $2$, then the conjugation $x\mapsto gxg^{-1}$ is an automorphims of $H$. As $C_5$ has only two automorphisms and the rest of the group structure is determined by these observations, there are no other groups ...


2

Hint: $G$ has an element $y$ of order $5$ and an element $x$ of order $2$. The sets $e, y, y^2, y^3, y^4$ and $ x, xy, xy^2, xy^3, xy^4$ are disjoint. Therefore these are all 10 elements in the group. Now $yx$ must be one of these elements. Show that $yx$ cannot be one of $e, y, y^2, y^3, y^4, x$.v So it can be $xy, xy^2, xy^3, xy^4$. If $yx=xy$ then ...


2

If $H$ is such a subgroup, then a Theorem of Frobenius tells us that there is $K \lhd G$ with $G = HK$ and $H \cap K = 1$. Furthermore, $K= \{1_{G} \} \cup (G \backslash \cup_{g \in G} g^{-1}Hg)$. Hence $|G| = |K| + [G:H](|H|-1).$ It follows that your seond inequality is violated precisely when $|K| \geq \frac{|G|}{2}.$ But $K$ is a subgroup of $G$, so the ...


2

You're using $^\times$ in correctly. $S^\times$ does not mean "remove zero". It means "remove all non units". In some contexts this is the same, because your set is a field, where every nonzero element is invertible. Definition: $x\in R$ is a unit (of $R$) if $xy=1_R$ has a solution. Another way of saying this is to ask for the invertible elements. ...


2

You will have to think a little more. Show that $g a g^{-1} \ne 1$, and then compute $(g a g^{-1})^{2}$. In $S_{3}$, take $a = (12)$ and $g = (123)$ and see what happens. $a g$ of order $2$ implies $a g a g = 1$. Now multiply on the left by $a$.


2

You can use the following fact. Of course, you should prove it first before using it (unless done in class). If $f:G\to A$ is a homomorphism of groups, and $A$ is abelian, then for all $x,y\in G$ their commutator $[x,y]=xyx^{-1}y^{-1}$ is mapped to the neutral element of $A$. Then I strongly advice you to compute $[i,j]$ in $Q$. This results in a ...


2

Hall's book "The Theory of Groups" proves (Theorem 12.5.4) that these groups, known as Hamiltonian groups, are all of the form $Q_8 \times A \times B$ where $A$ is an elementary $2$-group, $B$ is an abelian group where every element is of finite odd order. All groups of this form are Hamiltonian. Note there is no finiteness restriction on these $A$, $B$ ...


2

By Lagrange's theorem, the order of any element $g \in G$ must divide the order of $G$, which is odd. Therefore, the order of $g$ can't be even.


2

What you can say is that there is a surjective homomorphism $G \rtimes_\theta A\to G \rtimes_\phi Q$ given by $(g,a)\mapsto (g,\pi(a))$. Indeed, it is immediate from the definition of the semidirect product and the fact that $\theta=\phi\circ\pi$ that this map is a homomorphism, and it is obviously surjective. Furthermore, its kernel is the set of pairs ...


2

The operation is conjugation; $u^g = gug^{-1}$. The one of the definitions of a normal subgroup: $$N \trianglelefteq G \Leftrightarrow gNg^{-1} \subseteq N \Leftrightarrow gng^{-1} \in N\ \ \ \forall n \in N, g\in G$$ So as long as you verify that $U$ really is a subgroup, that test should tell you if it is normal or not. So you want to run something like ...


1

There is no need to find special counterexample. If $g^n = 1$ for all $g\in G$, then $g^{2n}=g^ng^n=1\cdot 1 = 1$ for all $g\in G$.


1

The smallest $m$ such that $g^m=1$ for all $g\in G$ is called the exponent of $G$. The exponent of $G$ is a divisor of the order of $G$ and in general is a proper divisor. For abelian groups, the exponent of $G$ equals the order of $G$ iff $G$ is cyclic.


1

Order of $3$ in $\mathbb Z_{10}^*$ is $4$, while the order of elements from $\mathbb Z_8^*$ is $2$.


1

Such a homomorphism is the same as a cyclic quotient of $Q$.


1

Unfortunately, we have a complete picture only up to around order of 2100. For the rest, we can calculate some, but these are either fixed and small, respectively smaller than 2100, or comparatively smaller than prime factors, or they are evidently much larger than $n$. Still, we can conjecture. And the conjecture says NO. I have created two graphs. Both ...


1

As noted by Geoff Robinson groups which have such a subgroup are called Frobenius groups, and in his answer he gave you the conditions when your seond inequality is true and when not. His cited results, namely that $K$ (the Frobenius kernel) is a subgroup is a rather deep result. I add a more elementary proof. Suppose $H \ne 1$ (otherwise the inequality is ...


1

The solution is correct, the number of permutations with a fixed length of the cycle containing 1 is $(n-1)!$. The length may be any number between $1$ and $n$, i.e. $n$ choices, so the total number of permutations is $n\times (n-1)!=n!$, as expected. Because this $(n-1)!$ is independent of $k$, we may say that all the lengths of the cycle including $1$ are ...


1

We can write $$c_\lambda =\sum_{g\in P_\lambda ,g^\prime\in Q_\lambda}sgn(g^\prime )\,e_{gg^\prime}$$ It is enough to show that coefficient of $e_{id}$ (where $id$ is the identity) in the above sum is non zero; in fact, we will show that the coefficient is $1$. This will follow from the fact that $gg^\prime =id$, for $g\in P_\lambda , g^\prime\in ...


1

Let $G=G_0$ be a finite group. Consider the set of proper nontrivial normal subgroups of $G_0$. If this set is empty, then $G_0$ is simple. Otherwise, the set is ordered by inclusion and we may choose a maximal element $G_1$. Note that by the correspondence theorem $G_0/G_1$ must be simple and we obtain a short exact sequence $$1\to G_1\to G_0\to ...


1

From the information on orbit lengths, you can deduce that $|G_\alpha|$ divides $12$. so we need to rule out $|G_\alpha|=12$ and $|A|=6$. I am only going to give a sketch proof. Suppose that $|A|=6$, so $A=\langle x,y \rangle$ with $o(x)=2$, $o(y)=3$. We know that $y$ fixes exactly $3$ points, so its centralizer in $R$ has order $3$, and we must have ...


1

$\mathbb{Z_{24}}$ is the only cyclic one - all other groups are non-abelian except $\mathbb{Z_{12}}\times \mathbb{Z_{2}}$, which you noted is not cyclic. Among the remaining ones, $\mathbb{Z_{12}}\times \mathbb{Z_{2}}$ is the only abelian one, so this is out too. If you have already seen the center of a group, you may proceed as follows. Among the ...



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