Let $D_8=\{a^ib^j:i\in\{0,1\},j\in\{0,...,3\}\} $ be a dihedral group, where $$a=\begin{pmatrix} -1 &amp;0 \\  0 &amp; 1 \end{pmatrix}\qquad\text{ and }\qquad b=\begin{pmatrix} cos\theta  &amp; -sin\theta \\  sin\theta &amp; cos\theta \end{pmatrix},$$ with $\theta=\frac{2\pi}{4}$. Prove that $D_8$ is not a Hamiltonian group.
122403
Let $H$ be the subgroup of the dihedral group $D_4$ generated by the elements $\left&lt;I\ (\text{identity}),p^2\epsilon\ (\text{flip across the horizontal diagonal})\right&gt;$. I have to tell whether or not this is a normal subgroup of the dihedral group $D_4$ (the regular polygon with 4 sides).   I know that $H = \{I,p^2\epsilon\}$. After that I made the multyplicative tale of D_4 and I saw that that $xH = Hx, \forall x \in D_4$, therefor H is normal in $D_4$. But I'm not sure if this is the way it should be done. I think that there is another way which does not imply the multyplicative table.
17295
I am wondering if my claim is correct, and how to formalize/write it? Thank you.     Give an example of a non-abelian group $G$ containing a proper normal subgroup $N$ such that $G/N$ is abelian.   So I am thinking of the dihedral group of triangle, with $3$ rotations and $2$ reflections. This is not abelian; but if I quotient $3$ rotations, the group has only two elements, which must be abelian.
3917
I'd like to know a reference for a simple proof that $\{c_\mu\mid \mu\vdash n\}$ is a basis for the centre of the symmetric group algebra $\mathbb{C}\mathfrak{S}_n$, where $c_\mu$ is the sum of all elements of a given cycle type (i.e. the sum of every element in the conjugacy class corresponding to $\mu$).
36721
Every undergraduate student knows that there are (exactly) two non-abelian groups of order 8: Dihedral ($D_8$) and Quaternion ($Q_8$). The group $Q_8$ has many interesting properties; simple of them are: it has unique element of order $2$, all subgroups are normal, etc. A property, which is strange and not obvious, the group $Q_8$ has, is that there is no group $G$ such that $G/Z(G)$ is isomorphic to $Q_8$. On the other hand, $D_8$ has no such property; there are (infinitely many) groups $G$ such that $G/Z(G)$ is isomorphic to $D_8$; example - $D_{16}$.   The question, I would like to ask, is related to a property of $D_8$, which is not shared by $Q_8$. There are (infinitely many) groups whose derived subgroup is $Q_8$; example- $SL_2(3)$. For groups $G$ of order &lt;100, one can check on GAP that $D_8$ is not a derived subgroup of any group. The natural question is then:  Question 1: Does there exist a finite/infinite group whose derived subgroup is equal to $D_8$?  Question 2: Does there exist a finite/infinite group whose derived subgroup is equal to (general) dihedral group $D_{2n}$ of order $2n$ ($n&gt;2$)?
71320
How do i show that $D_6$ has an element of order 6 but $A_4$ has only elements of order 3 and 2 and thus,  $A_4$ is not isomorphic to $D_6$? Can someone explain to me? Thanks!
49783
So I'm asked to prove that $$S_{3}\cong D_{3}$$ where $D_3$ is the dihedral group of order $6$. I know how to exhibit the isomorphism and verify every one of the $6^{2}$ pairs, but that seems so long and tedious, I'm not sure my fingers can withstand the brute force. Is there a better way?
71010
Let $G$ be the group  of all $2 \times 2$ matrices of the form $\begin{pmatrix} a &amp; b \\ 0 &amp; d\end{pmatrix}$ where $ad \neq 0$ under matrix multiplication. Let  $N=\left\{A \in G \; \colon \; A = \begin{pmatrix}1 &amp; b \\ 0 &amp; 1\end{pmatrix} \right\}$ be  a subset  of the group $G$. Prove that  $N$ is  a normal subgroup of $G$ and prove that  $G/N$ is abelian group.   Here is my attempt!  To prove $N$ is normal I consider  the group homomorphism $f \colon G \to \mathbb R^*$ given by $f(B) = \det(B)$ for all $B$ in $G$. And I see that $f(N)$ is all the singleton $\{1\}$ since $\{1\}$ as a subgroup of $\mathbb R^*$  is normal, it follows that $N$ is also normal. Is this proof helpful  here? Then how to prove that $G/N$ is Abelian? I know $G/N$ is a collection of left cosets.   Thank you.
83117
It's often said that there are only two nonabelian groups of order 8 up to isomorphism, one is the quaternion group, the other given by the relations $a^4=1$, $b^2=1$ and $bab^{-1}=a^3$.   I've never understood why these are the only two. Is there a reference or proof walkthrough on how to show any nonabelian group of order 8 is isomorphic to one of these?
84244
If $G$ is a group in which $(a\cdot b)^{i} =a^i\cdot b^i$ for three consecutive integers $i$ for all $a, b \in G$, show that $G$ is abelian.  Proof: Let $x$ be the smallest of the 3 consecutive integers. Then, we have   $(1)(a\cdot b)^{x} =a^x\cdot b^x$, $(2)(a\cdot b)^{x+1} =a^{x+1}\cdot b^{x+1}$ and $(3)(a\cdot b)^{x+2} =a^{x+2}\cdot b^{x+2}$.  Using $(2)$ and multiplying $a^{-1}$ on the left and $b^{-1}$ on the right, we get $baba...ba = a^{x}b^{x}(4)$ whereby there are $x$ number of $a$ and $x$ number of $b$ on both sides.  Using $(1)$ and multiplying both sides on the right by $ab$, we have $\overbrace{abab...ab}^{(x+1)ab} = (a^{x}b^{x})ab (5)$.  Substitute $(4)$ into $(5)$, we get $\overbrace{ab...ab}^{(x+1)ab} =\overbrace{ba...ba}^{(x)ba}ab(6)$.  Using $(3)$ and multiplying $a^{-1}$ on the left and $b^{-1}$ on the right, we get $\overbrace{ba...ba}^{(x+1)ba} = a^{x+1}b^{x+1} = \overbrace{abab...ab}^{(x+1)ab}(7)$.  Combining $(6)$ and $(7)$, and multiply $a^{-1}b^{-1}...a^{-1}b^{-1}$ on the left, we get $ab = ba$. Hence $G$ is abelian.
115756
Let $D_n$ be the dihedral group of order $2n&gt;4$; so it contains a cyclic subgroup $C$ of order $n$ on which $\sigma \in D_n$ outside of $C$ acts as $\sigma c \sigma^{-1}=c^{-1}$ for all $c \in C$. When is the cyclic subgroup $C$ with the above property unique? Determine all $n$ for which $D_n$ has a unique cyclic subgroup $C$ and justify your answer.  Isn't it true that all $D_n$ have a unique cyclic subgroup of order $n$, namely the one that includes all the rotations? And isn't it true that the reflections act the way described above on the rotations? But then the problem seems trivial, so perhaps I'm misunderstanding something?
44475
Let $n$ be any integer. We construct a group of order $n$ as  follows:   $G$ will consist of all symbols $a^i$, $i=0,1,2 \cdots, n-1$ where we   insist $a^0 = a^n = e$, $a^{i}\cdot a^{j} = a^{i+j}$ if $i+j \le n$   and $a^{i}\cdot a^{j} = a^{i+j-n}$ if $i+j &gt; n.$ The reader may verify   that this is a group. It is called a cyclic group of order $n$.   I'm trying to verify that this indeed is a group.  (a) For any $a^i, a^j \in G$ we have $a^i\cdot a^j = a^{i+j} \in G$ if $i+j \le n$; and $a^i\cdot a^j = a^{i+j-n} \in G$ if $i+j &gt; n$.  (b) For any $a^i, a^j, a^k \in G$, for $i+j+k \le n$ we have $a^i \cdot ( a^j \cdot a^k) = a^i \cdot a^{j+k} = a^{i+j+k}$ since $j+k \le n$ is implied by $i+j+k \le n.$ If $i+j+k &gt; n$ we have $a^i \cdot ( a^j \cdot a^k) = a^i \cdot a^{j+k}$ $= a^{i+j+k-n}$ if $j+k \le n$; and if $j+k &gt; n$ we have $a^i \cdot ( a^j \cdot a^k) = a^i \cdot a^{j+k-n}$ $= a^{i+j+k-2n}$.   For any $a^i, a^j, a^k \in G$, for $i+j+k \le n$ we have $(a^i \cdot  a^j) \cdot a^k = a^{i+j} \cdot a^{k} = a^{i+j+k}$ since $i+j \le n$ is implied by $i+j+k \le n.$ If $i+j+k &gt; n$ we have $(a^i \cdot  a^j) \cdot a^k = a^{i+j} \cdot a^{k}$ $= a^{i+j+k-n}$ if $i+j \le n$; and if $i+j &gt; n$ we have $(a^i \cdot  a^j) \cdot a^k = a^{j+k-n} \cdot a^{k}$ $= a^{i+j+k-2n}$.   (c) There's $a^0 \in G$ such that for any $a^i \in G$ we have $a^i \cdot a^0 = a^{i+0} = a^i$ and $a^0a^i = a^{0+i} = a^i.$  (d) For every $a^i \in G$ there's $a^{-i} \in G$ such that $a^i \cdot a^{-i} = a^{i-i} = a^0$ and $ a^{-i} \cdot a^i  = a^{-i+i} = a^0.$  Is this correct?
111109
Let $G$ be a group. Prove or disprove that $H =\{g^2 | g \in G\}$ is a subgroup of $G$.   I tried testing the permutations of $A_4$, however squaring each cycle yielded a cycle in $A_4$ so I'm lacking a counter-example (if there is one). In a nutshell I'm looking for a subgroup such that when you square the permutation cycle, it yields a cycle not in that subgroup.  Or, I could be way off base and figure out that there isn't a counter-example and I need to prove that indeed $H$ is a subgroup of $G$.
120173
Prove that $D_{12}\cong S_3 \times C_2$.   I really dont know how I should start this question. My gut feeling says in some way I have to consider normal subgroups of $D_{12}$ but I cannot see how this will lead necessarily to a unique solutions.  No full solutions please hints only (partly because I cannot give any more of an attempt and I dont want this downvoted)  Ideally I would like a non geometric solution so that similar techniques can be used for general groups
93809
I need some help with the following task: Show that  for all $n\geq2$ is $D_{2n}/Z(D_{2n})\cong D_n$ $D_n$ is the dihedrial group with $2n$ elements and looks like $D_n=\{id,r,...,r^{n-1},s,sr,...,sr^{n-1}\}$ Sadly I have no idea how $D_{2n}/Z(D_n)$  looks like but I guess I need to know that to show that it is a homomorphism and bijectiv. Thanks for helping me
110333
Let $G$ be a group, and let $n\in\mathbb Z$. Show that $(ab)^n=a^nb^n\iff(ab)^{1-n}=a^{1-n}b^{1-n}.$   For the $"\implies"$ direction, let's assume $(ab)^n=a^nb^n$, and let's consider the inverse of $ab$. We can write $(ab)^{-n}=b^{-n}a^{-n}$. If we multiply by $ab$, we get $$ (ab)^{1-n}=(ab)(ab)^{-n}=(ab)b^{-n}a^{-n}=ab^{1-n}a^{-n}. $$  But now I need commutativity to conclude the proof. Could someone help me with the last step?
35799
Given an example of a non-abelian group $G$ satisfying $(a \cdot b)^i=a^i \cdot b^i, \forall a, b \in G$ for two consecutive integers.   This is question 5 from Herstein Page 35. I have proved that conclusion holds for three positive integers.Thanks.
103727
Question is to prove that :     $Q_8$ is isomorphic to a subgroup of $S_8$, but not isomorphic to a subgroup of $S_n$ for $n\leq 7$.   I see that $Q_8$ is isomorphic to subgroup of $S_8$ by left multiplication action.  Hint given was to prove that stabilizer of any point contains $\{\pm 1\}$.  To prove Cayley's theorem, stating any group is isomorphic to a subgroup of $S_n$ we take action of given group on a set $A$ having same cardinality.  with that motivation I want to check if there is a Isomorphism then there is a map from $G\times A \rightarrow A$. i.e., $G$ gives a permutation group $S_A$.  I tried in same manner. Suppose $Q_8$ is isomorphic to subgroup of $S_n$ with $n\leq 7$  then it should come from a group action of $Q_8$ on a set of cardinality atmost 7.   Suppose $Q_8$ acts on a set $A$ with possible cardinality at most 7.  $stab(a)=\{g\in Q_8 : g.a=g \forall a\in A \}$  $cl(a)=\{g.a : g\in Q_8\}$  I know number of elements in class of $a$ equals to index of stabilizer.  As $cl(a)=\{g.a : g\in Q_8\}\in A$ i.e., $cl(a)\subseteq A$ and as $|A|\leq 7$ i see that $|cl(a)|\leq 7$.  But, $|cl(a)|=|Q_8:stab(a)|$ for any element $a\in A$.  So, $|Q_8:stab(a)|=|cl(a)|\leq 7$ for all $a\in A$.   So, $stab(a)$ should be non trivial subgroup of $Q_8$ if not then $|Q_8:stab(a)|=8$  non trivial subgroup (proper) of $Q_8$ contains $\{\pm1\}$.  So, In the worst case, $\{\pm 1\}\subseteq stab(a)$ for all $a\in A$.  As $Ker(\eta)=\cap_{a\in A}stab(a)$ (where $\eta$ is the action of $Q_8$ on $A$)   we see that $\{\pm 1\}\subseteq Ket(\eta)$ which means that $Ker(\eta)$ is non trivial.  thus there is no isomorphism (coming from $\eta$) between $Q_8$ and any subgroup of $S_7$.  I would be thankful if someone can check whether my approach is correct or if there is any other simple possible way.  P.S : Usually what i do to see whether two groups are isomorphic or not is to check for cardinality, abelian property, no of elements with same order and so on. But I was having no idea when i fail in all these ways. With this Group actions i could see possibility for getting a precise conclusion on Isomorphisms.I would like to Thank Mr. Jyrki Lahtonen (a user of Math.SE) who made me to get used to Group actions.  P.S $2$: If any thing is wrong in my idea, it is entirely my fault, and if anything is correct in this whole credit should go to Mr. Jyrki Lahtonen