im trying to show that $D_6  \cong D_3\times \mathbb Z_2$ as groups.  $D_n$ is the is the dihedral of order 2n  I know that it is possible to form all of the points in $D_6$ and in $D_3$ as compositions of $R$ the rotation operation (such that every vertex move one to the left) and T the reflection on one of the symmetry axis, so i tried to write it done and joining point that act the same, but for no use.
22095
Let $n \geq 1$ be an odd integer. Show that $D_{2n}\cong \mathbb{Z}_2 \times D_n$. I define a map $$\phi:D_{2n} \rightarrow \mathbb{Z}_2 \times D_n$$ by $R \mapsto (0,r^{\frac{n+1}{2}})$ and $M \mapsto(1,m)$. Then I am stuck at showing the map is bijective. By the way, do we need to show that $(0,r^{\frac{n+1}{2}})(1,m)(0,r^{\frac{n+1}{2}})^{-1}=(1,m)$ ? If we can prove this, then what can we conclude?
29223
I'm a bit lost on this problem:      Show that a group of order $p^2$ is isomorphic to $\mathbb{Z}_{p^2}$ or $\mathbb{Z}_{p}$ $\times$ $\mathbb{Z}_{p}$.    The only thing I can think of that may relate to this problem is Lagrange's Theorem, where the order of a subgroup divides the order of a group. But I'm not sure if I can even use that for this problem... Any help would be tremendously helpful. Thank you!
45636
Let $Q_8$ be the quaternion group. Show that $G/Z$ can never be isomorphic to $Q_8$, where $Z$ is the center of $G$. Hint: if $G/Z\cong Q_8$, show that $G$ has two abelian subgroups of index $2$.  I'm trying to prove the hint by looking at the canonical homomorphism $\pi:G\rightarrow G/Z$. Well, $Q_8$ has some abelian subgroups of index $2$, such as $H=\{1,-1,i,-i\}$, but that doesn't seem to map to abelian subgroups of index $2$ of $G$. (We know that $\pi(ab)=\pi(a)\pi(b)=\pi(b)\pi(a)=\pi(ba)$ for $a,b$ in $\pi^{-1}(H)$, but that only yields $aba^{-1}b^{-1}\in Z$, not $aba^{-1}b^{-1}=1$.)
98793
Question is to Prove that $(\mathbb{Z}/2^n \mathbb{Z})^*$ is not  cyclic Group for $n\geq 3$.  Hint : Find two subgroups of order $2$.  I somehow feel that a cyclic group can not have two distinct groups of same order. but, I am not sure about the proof.  I have no idea how to proceed for this.   any hint would be appreciated.  Thank You.
75557
The group of orthogonal transformations of $\mathbb{R}^3$ is direct product of the group of rotations and the group $Z=\langle x\mapsto -x\rangle\cong \mathbb{Z}/2$. The finite subgroups of group of rotations are well known: rotational symmetries of regular polyhedrons.   These subgroups, along with their direct product with $Z$ will give some (families) of finite groups of orthogonal transformations.  It looks that some finite groups of orthogonal transformations need not be of this form (see Shafarevich- Algebra I; I couldn't understand arguments there).  How do we cover all finite subgroups of orthogonal transformations?  A more general question in this context may be: How to get subgroups of $H\times K$ knowing subgroups of $H$ and $K$?
36106
Is $D_n$, the dihedral group of order $2n$, isomorphic to a subgroup of $S_n$ ( symmetric group of $n$ letters) for all $n&gt;2$?
87787
I am trying to classify groups of order 18. So far, I have shown that a group $G$ of order 18 is given by $G\cong C_9 \rtimes_{\varphi} C_2$ or $G\cong (C_3 \times C_3)\rtimes_{\varphi} C_2$.  If $G\cong C_9 \rtimes_{\varphi} C_2$, then $\mid \varphi(1) \mid$ divides $2$, so that $\varphi(1)$ is trivial or inverts a generator of $C_9$. I concluded that $G \cong C_{18}$ in the former case and $G \cong D_{18}$ in the latter case.  I am now considering the case $G\cong (C_3 \times C_3)\rtimes_{\varphi} C_2$, but I am stuck. I have found , but they do not do it the same way as me (finding the image of $\varphi(1)$ in $Aut(C_3 \times C_3)$).  Is there a way to do this using my method? My goal is to be able to do questions like this on my algebra qual without having to fiddle around. I was messing around with the fact that $Aut(C_3 \times C_3)\cong GL_2(C_3)$.
118408
$D_3=\big&lt;a,b| a^3=b^2=(ab)^2=1\big&gt;$ and $|D_3|=6  $   $D_3=\{  1, a, a^2, b, ab, a^2b\} $  Are these correct?  Then so that I can compute the orders of all the elements of the group $D_3$
46555
I am trying to solve the following exercise about the dihedral group and its center:  If $g\in Z(D_{2n})\Leftrightarrow ga=ag, bg=gb$, where $a,b$ are generators of $D_{2n}$.  We have defined the dihedral group of order $2n$ as $D_{2n}=\left \{ a^{i}b^{j} \mid 0\leq i \leq n, j=0,1 \right \}$, the group of all finite products of the generators $a$ and $b$,  where  $a= \bigl(\begin{smallmatrix}     1 &amp; 2 &amp; 3 &amp; \cdots &amp; n-1 &amp; n \\     \end{smallmatrix}\bigr)$ and $ b = \bigl(\begin{smallmatrix}     1 &amp; 2 &amp; 3 &amp; \cdots &amp; n-1 &amp; n \\     1 &amp; n &amp; n-1 &amp; \cdots &amp;  3  &amp; 2   \end{smallmatrix}\bigr)$  For the first part ( $\Rightarrow $) i have considered an element $g$ of the center $Z(D_{2n})$, $g$ is also in $D_{2n}$, so it can be written as a finite product of powers of $a$ and $b$. Let's say $g=a^{k}b^{l}$ for $0 \leq k \leq n$ and $j=0$ or $j=1$.  Then i can write $ga = a^{k}b^{l}a=a^{k+1}b^{l}$ and $ag=aa^{k}b^{l}=a^{k+1}b^{l}$. So $ga=ag$. The same for $gb=bg$.   Is this correct? How to start with the other part of the proof? I would appreciate any help and hints. Thank you in advance.
26097
I have a few questions concerning an example of the commutator subgroups in the dihedral group. This example is found on pg.171 of Abstract Algebra by Dummit and Foote.      Let $D_{2n}=\langle r,s |r^n=s^2=1, s^{-1}rs=r^{-1}\rangle$. Since $[r,s]=r^{-2}$ we have that $\langle r^{-2} \rangle = \langle r^2 \rangle \le D'_{2n}$. Furthermore, $\langle r^2\rangle \trianglelefteq D_{2n}$ and the images of $r$ and $s$ in $D_{2n} / \langle r^2 \rangle$ generate this quotient.    What exactly is meant by the image of $r$ and $s$?     They (is this referring to $r$ and $s$?) are commuting elements of order $\le 2$ (I know $s$ is of order $2$ but $r$ is of order $n$??) so the quotient is abelian and $D'_{2n} \le \langle r^2 \rangle$. (I thought the quotient was abelian due to the already established properties of $\langle r^2 \rangle$ i.e it is normal and a subgroup of the commutator subgroup.)
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?
48429
I am completely stuck at the question     Let $G$ be a finite group and let $x$ and $y$ be distinct elements of order 2 in $G$ that generate $G$. Prove that $G \cong D_{2n}$, where $n = |xy|.$   I have proved that     Let $x$ and $y$ be elements of order 2 in any group $G$. If $t = xy$ then $tx = xt^{-1}$.   Can I get some hints?
37662
List all the conjugate classes in the dihedral group of order $2n$ and verify the class equation.  The dihedral group is generated by two elements $r$ and $s$.   The order of $r$ is two since $r^2=e$ and $s$ is $n$ since $s^n = e$. And I know all elements can be produced as either $s^k$ or $rs^k$. How can I list all the conjugacy class?
20705
In general, the group $\mathbb Z_{mn}$ under addition $\bmod mn$ is isomorphic to the direct product $\mathbb Z_m \times \mathbb Z_n$ if and only if $\gcd(m,n)=1.$ This is easy to see since $\mathbb Z_{mn}$ is generated by the element $(1,1)$, and if $\gcd(m,n)=1$, then this will only be zero if it is multiplied by itself $mn$ times (resulting in the cyclic group).   However I'm interested in constructing an isomorphism between the two groups as an explicit expression. I've been playing around a bit with some examples, one of which is the function $\phi : \mathbb Z_2 \times \mathbb Z_3 \to \mathbb Z_6$ where $\phi(a,b) = 3a + 4b \pmod 6$, which gives us the desired isomorphism. I'm interested here in the relationship between the integers $2$, $3$, and the coefficients $3, 4$ of $a$ and $b$. After playing around in Wolfram|Mathematica, I managed to obtain the following isomorphisms $\phi_{m,n} : \mathbb Z_m \times \mathbb Z_n \to \mathbb Z_{mn}$ for coprime $m,n$ in $\{1, 2, \dots, 10\}$.   First of all, the following results seem to hold in general.  $$\phi_{1, n}(a, b) = a + b \qquad \text{for all $n$}$$  $$\phi_{m, 1}(a, b) = a + b \qquad \text{for all $m$}$$  $$\phi_{2, n}(a, b) = na + (n + 1)b \qquad \text{for all $n$}$$ $$\phi_{m, n}(a, b) = K a + L b \iff \phi_{n,m}(a, b) = L a + K b \qquad \text{for all $m, n$}$$  Beyond this point, I could not spot a general pattern. The isomorphisms I obtained are:  $$\phi_{3, 4}(a, b) = 4a + 9b$$ $$\phi_{3, 5}(a, b) = 10a + 6b$$ $$\phi_{3, 7}(a, b) = 7a + 15b$$ $$\phi_{3, 8}(a, b) = 16a + 9b$$ $$\phi_{3, 10}(a, b) = 10a + 21b$$ $$\phi_{4, 5}(a, b) = 5a + 16b$$ $$\phi_{4, 7}(a, b) = 21a + 8b$$ $$\phi_{4, 9}(a, b) = 9a + 28b$$ $$\phi_{5, 6}(a, b) = 6a + 25b$$ $$\phi_{5, 7}(a, b) = 21a + 15b$$ $$\phi_{5, 8}(a, b) = 16a + 25b$$ $$\phi_{5, 9}(a, b) = 36a + 10b$$ $$\phi_{6, 7}(a, b) = 7a + 36b$$ $$\phi_{7, 8}(a, b) = 8a + 49b$$ $$\phi_{7, 9}(a, b) = 36a + 28b$$ $$\phi_{7, 10}(a, b) = 50a + 21b$$ $$\phi_{8, 9}(a, b) = 9a + 64b$$ $$\phi_{9, 10}(a, b) = 10a + 81b$$ If you want to try and generate a few yourself, here is the Mathematica function I implemented. I'm sure it can be optimised, but for the first 10 coprime integers it sufficed.  isomorphism[{m_, n_}] :=       Module[{x = Null},       For[i = 1, i &lt; m*n,           i++, (If[#[[2]] == Table[i, {i, 0, m*n - 1}],           x = ToString[#[[1]][[1]]] &lt;&gt; "a+" &lt;&gt; ToString[#[[1]][[2]]] &lt;&gt;          "b"; Break[]] &amp;) /@       Table[{{i, j},       Mod[i #[[1]] + j #[[2]], m*n] &amp; /@       Table[{Mod[i, m], Mod[i, n]}, {i, 0, m*n - 1}]}, {i, 1,      m*n}, {j, 1, m*n}][[i]]]; x];   I'd appreciate if anyone can help me find the pattern to determine $\phi_{m, n}$ in general.
68191
Consider the dihedral group  $$D_{2n}= \langle a,b \mid a^n = 1 = b^2, b^{-1}ab = a^{-1}\rangle$$  How can I show that $|D_{2n}| = 2n$?   I'm trying to show that we can write every element in the form  $$x = a^i b^j$$ where $i= 0, 1, \dots, n-1$; $b = 0,1$.   I managed to show existence, and it is clear that there are $2n$ such elements, so if I can show that every choice of $i,j$ gives a distinct element, I'll be done.  Any ideas?
62086
The exercise for which I am seeking feedback wants me to prove that the center of a direct product is the direct product of the centers. That is, $$Z(G_1 \times \cdots \times G_n)=Z(G_1) \times \cdots \times Z(G_n).$$  $\textbf{Attempt at Solution:}$ We proceed by induction on $n$. First, we prove our desired for $n=2$; this will establish our basis step. Let $(g_{1},g_{2}) \in Z(G_{1} \times G_{2})$. Then $(g_{1},g_{2})(x_{1},x_{2})=(x_{1},x_{2})(g_{1},g_{2})$ for all $(x_{1},x_{2}) \in G_{1} \times G_{2}.$ This implies $g_{1}x_{1}=x_{1}g_{1}$ and $g_{2}x_{2}=x_{2}g_{2}$ for all $x_{1} \in G_{1}$ and for all $x_{2} \in G_{2}$. Hence, $g_{1} \in Z(G_{1})$ and $g_{2} \in Z(G_{2})$. To prove the reverse inclusion, this time let $(g_{1},g_{2}) \in Z(G_{1}) \times Z(G_{2})$ Then, $g_{1}x_{1}=x_{1}g_{1}$ and $g_{2}x_{2}=x_{2}g_{2}$ for all $x_{1} \in G_{1}$ and for all $x_{2} \in G_{2}$; it, of course, follows that $(g_{1},g_{2})(x_{1},x_{2})=(x_{1},x_{2})(g_{1},g_{2})$ for all $(x_{1},x_{2}) \in G_{1} \times G_{2}.$ Indeed, $(g_{1},g_{2}) \in Z(G_{1} \times G_{2})$  For our induction step, assume that desired result holds for some $n$; that is, assume $$Z(G_1 \times \cdots \times G_n)=Z(G_1) \times \cdots \times Z(G_n).$$ We show $$Z(G_1 \times \cdots \times G_n \times G_{n+1})=Z(G_1) \times \cdots \times Z(G_n) \times Z(G_{n+1}).$$ By our basis step, $$Z(G_1 \times \cdots \times G_n \times G_{n+1})=Z(G_1 \times \cdots \times G_n) \times Z(G_{n+1}).$$ Then, using our induction hypothesis, we obtain $$Z(G_1 \times \cdots \times G_n) \times Z(G_{n+1})=Z(G_1) \times \cdots \times Z(G_n) \times Z(G_{n+1}).$$ Hence, our induction step has been established. Finally, by the principle of mathematical induction, we conclude $$Z(G_1 \times \cdots \times G_n)=Z(G_1) \times \cdots \times Z(G_n),$$ for any finite collection of groups $G_{1}, \ldots , G_{n}.$  $\blacksquare$  $\textbf{Question:}$ This may be a stupid question; but, I'll ask anyway! Would it be wrong if I completed the proof by letting $(g_{1},\ldots , g_{n}) \in Z(G_1 \times \cdots \times G_n)$ and showing $(g_{1},\ldots , g_{n}) \in Z(G_1) \times \cdots \times Z(G_n)$? (Then, of course, proceed by proving the reverse inclusion.) Thank you!
77584
We need to prove that $D_{3} \oplus D_{4}$ is not isomorphic to $D_{24}$ .  The way in which I approach such type of questions is to count the number of elements of order $x$ in one group and then in the other group, and then conclude that they aren't equal and hence there can't be any isomorphism between them.  But this approach here doesn't look easy, with $D_{24}$ involved.  Could anyone suggest another way of solving this ?