I have written that all the groups with 4 element are isomorphic with Z4 or the Klein group. Can somebody explain me why,please?
104048
I will try and show that if our group $G$ is not cyclic then it is isomorphic to .  So suppose $G$ is not cyclic. Then its elements, excluding $e$, are of order $3$ or $2$. Supposing there is an element of order $3$ gives three distinct elements $g, g^2, g^3$. By hypothesis there exists  $h \in G$ such that $h$ is distinct from these three elements. So we can make $gh$ which must be equal to one of the elements previously mentioned. Setting this equal to the various elements:  \begin{align} &amp;gh=g \Rightarrow h=e &amp;&amp; gh=g^2 \Rightarrow h=g &amp; \\ &amp;gh=h \Rightarrow g=e, &amp;&amp; gh=g^3 \Rightarrow h=g^2 &amp; \end{align} Each of which is a contradiction. Therefore all the elements are of order two and we can get an isomorphism between $G$ and $H \times H$ where $ H$ is cyclic of order two. But this direct product is isomorphic to Klein's four-group so $G$ is isomorphic to $V$.  Is this correct?  Also it is just given in the book that $V$ is isomorphic to this product; how do I show this without checking that $f(gh)=f(g)f(h)$ for all combinations?
51827
Given that $C_n$ is a cyclic group of order $n$, what conditions must integers $n$ and $m$ satisfy such that the group $C_n \times C_m$ is isomorphic to C$_{mn}$?   So I attempted to investigate a few lower-order cyclic groups to find a pattern. In particular, I looked at $C_2 \times C_3$ and compared it to $C_6$. Here are a couple of diagrams:  $C_2 \times C_3$ -   $C_6$ -   They Cayley tables look pretty different (identity elements are all over the place for $C_2 \times C_3$) so I concluded that they are not isomorphic.  I did the same for $C_2 \times C_4$ and $C_8$:  $C_2 \times C_4$ -   $C_8$ -   Again, I arrived at the conclusion that the 2 groups are not isomorphic due to their Cayley table structure.  It seems I'm stuck now. Are my conclusions wrong for the examples I examined? If so, how do I find the conditions for the general case of comparing $C_m \times C_n$ with $C_{mn}$?
7537
I am trying to prove that the quotient group $SL_2(\mathbb Z_3)/Z(SL_2(\mathbb Z_3))$ is isomorphic to $A_4$. I could show that $SL_2(\mathbb Z_3)$ has $24$ elements (one can see this calculating all the elements by hand or simply by noticing that $GL_2(\mathbb Z_3)$ has $48$ elements and that $SL_2(\mathbb Z_3)$ is the kernel of the morphism $\det:GL_2(\mathbb Z_3) \to G_2$). The center of $SL_2(\mathbb Z_3)$ is $\left\{\begin{pmatrix}1&amp; 0\\0&amp; 1\end{pmatrix}, \begin{pmatrix}-1&amp; 0\\0&amp; -1\end{pmatrix}\right\}$. I got stuck trying to show these two groups are isomorphic. By Sylow theory I can say that $n_3=4$, where $n_p=$ number of $p$-Sylow subgroups. Then I can define the action by conjugation on $X$ the set of all $3$-Sylow subgroups. So I have a morphism $\phi:SL_2(\mathbb Z_3) \to S(X)$. It is clear that $Z(SL_2(\mathbb Z_3)) \subset \ker \phi$. I don't know what to do from here. I would appreciate suggestions or an alternative solution to this problem. Thanks in advance.
11503
Hello the question is as stated above and is given to us in the context of group theory, specifically under the heading of isomorphism and products. I would write down what I have tried so far but I have made very little progress in trying to solve this over the last few hours!
49957
I'm trying to find 4 groups of order 12, none of which are isomorphic to each other.  Should I be trying external direct products?   So far I have $A_4, \mathbb Z_{12},\,$ and $\,\mathbb Z_6\times \mathbb Z_2.\,$ How do I show all of these are non-isomorphic to each other? And how do I find a fourth? `
122369
Prove that $\mathbb{Z}$ is not isomorphic to additive group of any vector space over any field.  Proof.  Surpose that: $\phi : (A, +) \rightarrow  \mathbb{Z} $ is an isomorphism. Then there is some $r∈A $ such that $ϕ(r)=1_\mathbb{Z}$. We have $$ 1_\mathbb{Z}=ϕ(r)=ϕ(2(r/2))=2ϕ(r/2)$$  or equivalently that $ϕ(r/2)=\frac{1}{2}$. But this is not in $ \mathbb{Z}$, so there can be no such morphism.  Is my proof above correct? Thanks
123633
I was thinking about the following problem:  Suppose that $G_1 \cong G_2$ are isomorphic groups. Under what conditions on the groups $H_1,H_2$ will we have $$G_1 \times H_1 \cong G_2 \times H_2 ?$$  Obviously, in the finite case we must have $|H_1|=|H_2|$. Also, it is easy to see that $H_1 \cong H_2$ is sufficient. Is it necessary as well?  Thank you!
97344
Prove that $S_3\times \mathbb{Z}_2$ is isomorphic to $D_6$.  My attempt:  I tried to decompose $D_6$ as the internal direct product of subgroups $H$ and $K$ isomorphic to $S_3$ and $\mathbb{Z}_2$ respectively:  Let $H=\{1, r^2, r^4, s, r^2s, r^4s\}$  Let $K=\{1, r^3s\}$  $HK=D_6$  $H\cap K=\{1\}$  However, taking $h=r^2$ and $k=r^3s$, \begin{equation*} \begin{split} hk&amp;=r^2\cdot r^3s \\ &amp;=r^5s \end{split} \end{equation*} \begin{equation*} \begin{split} kh&amp;=r^3s\cdot r^2 \\ &amp;=r^3sr^2ss \\ &amp;=r^3r^{-2}s \\ &amp;= rs \end{split} \end{equation*} $hk\neq kh$  So, this doesn't work. How do I solve this problem?
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!
20011
Let $H=\{e,(1 2)(3 4)\}$ and $K=\{e,(12)(34),(13)(24),(14)(23)\}$ be subgroups of $S_4$, where $e$ denotes the identity element of $S_4$. Which of the following are correct?   $H$ and $K$ are normal subgroups of $S_4$. $H$ is normal in $K$ and $K$ is normal in $A_4$. $H$ is normal in $A_4$ but not normal in $S_4$. $K$ is normal in $S_4$ but $H$ is not.  I know the condition for a subgroup $H$ to be normal in $G$ but that includes too much computation as $S_4$ has $24$ elements in all. I have studied the group of symmetries of square i.e. $D_4$ and its geometrical interpretation and I think that might be a clue to solve such type of problem relatively easy. Kindly correct me if I am wrong and provide any possible short method to do this.
18492
Let $S_4$ be a symmetric group on $4$ elements, $V_4$ - its subgroup, consisting of $e,(12)(34),(13)(24)$ and $(14)(23)$ (Klein four-group). $V_4$ is normal and $S_4/V_4$ if consisting of $24/4=6$ elements. Hence $S_4/V_4$ is cyclic group $C_6$ or a symmetric group $S_3$ (really, there are only two groups consisting of $6$ elements). It is easy to see, that an order of each element of $S_4$ is $1,2,3$ or $4$. So, $S_4/V_4$ is isomorphic to $S_3$.  My question: how to build the isomorphism explicitly?
26441
A question that may sound very trivial, apologies beforehand. I am wondering why $( \mathbb{Q}_{&gt;0} , \times )$ is not isomorphic to $( \mathbb{Q} , + )$. I can see for the case when $( \mathbb{Q} , \times )$, not required to be positive, one can argue the group contains elements with order 2 (namely all negatives). In the case of the requirement for all rationals to be positive this argument does not fly. What trivial fact am I missing here?
46731
I am stuck in the following problem:  Show that every group of order 4 is an extension of $\mathbb{Z}_{2}$ by $\mathbb{Z}_{2}$. Which of the exact sequences splits?  Ok, i have to consider two cases for a group of order 4: $\mathbb{Z}_{4}$ and $\mathbb{Z}_{2}\times \mathbb{Z}_{2}$, so i get two exact sequences:  $0\rightarrow \mathbb{Z}_{2}\overset{\psi }{\rightarrow}\mathbb{Z}_{4}\overset{\varphi }{\rightarrow}\mathbb{Z}_{2}\rightarrow 0$  and $0\rightarrow \mathbb{Z}_{2}\overset{\tilde{\psi} }{\rightarrow}\mathbb{Z}_{2}\times \mathbb{Z}_{2}\overset{\tilde{\varphi} }{\rightarrow}\mathbb{Z}_{2}\rightarrow 0$  For both cases i checked the conditions of exact sequence and proved that these are exact sequences. Now i have a problem how to show if they split or not.  We say that an exact sequence splits if for the first sequence there exists a group homomorphism $\pi :\mathbb{Z}_{2}\rightarrow \mathbb{Z}_{4}$ with $\varphi \circ \pi =id_{\mathbb{Z}_{2}}$ and for the second $\tilde{\varphi} \circ \tilde{\pi} =id_{\mathbb{Z}_{2}}$, right? But how to show if these group homomorphisms exist or not?  Can anybody help me, please? I would appreciate any hints and comments. Thank you in advance!
107359
Prove $D_{8n} \not\cong  D_{4n} \times Z_2$.   My trial:  I tried to show that $D_{16}$ is not isomorphic to $D_8 \times Z_2$ by making a contradiction as follows:   Suppose $D_{4n}$ is isomorphic to $D_{2n} \times Z_2$, so $D_{8}$ is isomorphic to $D_{4} \times Z_2$. If $D_{16}$ is isomorphic to $D_{8} \times Z_2 $, then $D_{16}$ is isomorphic to $D_{4} \times Z_2 \times Z_2 $, but there is not Dihedral group of order $4$ so $D_4$ is not a group and so $D_{16}\not\cong D_8\times Z_2$, which gives us a contradiction. Hence, $D_{16}$ is not  isomorphic to $D_{8} \times Z_2$.   I found a counterexample for the statement, so it's not true in general, or at least it's not true in this case.  __  Does this proof make sense or is it mathematically wrong?
117005
Are the groups $\mathbb{C}$ and $\mathbb{R}$ isomorphic under addition?  And how could I prove this ?  What about $\mathbb{Q}$ and $\mathbb{Q}[i]$ ?
108508
How to show $D_3\oplus D_4$ is not isomorphic to $D_{24}$?  Here $D_n$ is the dihedral group of order $2n$.  I am not sure how to prove this. I am not very good with the dihedreal groups.
120185
states that the isomorphism type of a finite simple group is determined by its order, except that:   $L_4(2)$ and $L_3(4)$ both have order $20160$ $O_{2n+1}(q)$ and $S_{2n}(q)$ have the same order for $q$ odd, $n &gt; 2$   I think this means that for each integer $g$, there are $0$, $1$ or $2$ simple groups of order $g$.  Do we need the full strength of the  to prove this, or is there a simpler way of proving it?