I know that ${\rm Aut} (\Bbb Z/p \times \Bbb Z/p)$ is $2\times 2$ invertible matrix with all possible entries mod $p$. I was asked to count how many elements in this group, but it gets complicated as I tried to count by dividing into several cases. How can I go about it?
121857
I happen to know that Aut($\mathbb{Z_p} \times \mathbb{Z_p}$) $\simeq$ $GL_2(p)$, and so has order $p(p^2-1)(p-1)$.  Which means something is wrong with the following argument:  An automorphism $\phi$ is determined by its value on the two generating elements $(1,0) $ and $(0,1)$ of $\mathbb{Z_p} \times \mathbb{Z_p}$.  These must be sent to elements of the same order, $p$. In fact every nonzero element of $\mathbb{Z_p} \times \mathbb{Z_p}$ has order $p$, so there are $p^2-1$ choices to which each generator must go. Therefore the number of automorphisms is $(p^2-1)^2$.  So I've overcounted by $(p^2-1)(p-1)$; I imagine some of the homomorphisms in my argument were not distinct, or not isomorphisms. is there a way to easily rectify for the overcounting? Or do we throw the whole argument away?
67836
I am attempting to understand and compute: $(\mathbb{Z}_4 \times \mathbb{Z}_6)/\langle(0,2)\rangle$   I know $(0,2)$ generates $H = \{(0,0),(0,2),(0,4)\}$, which has  order 3 because there are 3 elements. Now, I must find all the cosets  which there should be 8. The reason there should be 8 is because $4\cdot 6/3 = 8$.  This is where I run into the issue. I have a hard time generating the cosets, but furthermore, once I actually do get the cosets, I have a hard time analyzing them and knowing what it means. If someone could walk me through a step-by-step solution that would be great.  The answer is that the factor group is isomorphic to $\mathbb{Z}_4 \times \mathbb{Z}_2$.   Reference: Fraleigh, A First  Course in Abstract Algebra, p. 147, Example 15.10.
59108
Let $𝐺$ be the group generated by the matrices $𝐴=\begin{bmatrix}2&amp;0\\0&amp;1\end{bmatrix}$, $𝐵=\begin{bmatrix}1&amp;1\\0&amp;1\end{bmatrix}$, where the group operation is matrix multiplication.      Prove that $𝐺$ contains a subgroup $𝐻_𝑖$ for each $𝑖\in \mathbb{Z}$ such that   $$ \cdots \subset H_{-2}\subset H_{-1} \subset H_0 \subset H_1 \subset \cdots$$   So this is all that is given and i'm not sure where to start from this question. I know that $ABA^{-1}$ = $B^{2}$. From this we can replace A with $A^2$ and so on but i am not sure how i would prove this. Can someone help please?
61948
I am trying to show that if for a group $G$ we have $Z(G)=1$, then $Z(\operatorname{Aut}(G))=1$.  I have tried everything I can think of, but to no avail.  The first part of the problem was to show that $\operatorname{Inn}(G) \triangleleft  \operatorname{Aut}(G)$ - does this have any connection?  I couldn't find one.  I know that $\operatorname{Inn}(G) \cong G/Z(G)$, so in this case $\operatorname{Inn}(G) \cong G$ and $Z(\operatorname{Inn}(G))=1$.  Is there any way to get from this to $Z(\operatorname{Aut}(G))=1$?  How do we go about showing this?  Thanks.
38667
Show that number of elements in $(Z/nZ)^*$ is $\phi (n)$. $(Z/nZ)^*$ is set having those classes whose inverse exists.  If we prove that $(Z/nZ)$ has elements whose inverse exists are those who are relatively prime to n. Then my question will be proved  So i write let $gcd(a,n)=1$  $1= as+nt$   $as=1mod(n)$  as ~ 1  so class of a . class of s = 1  i.e inverse of class of a exists  Kindly suggest is this correct or some alternative option  Thanks
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.
123173
The set $Z_n = \{ 0, 1, . . . , n-1 \}$ for $n ≥ 1$ is a group under addition modulo n. For any $j &gt; 0$ in $Z_n$, the inverse of $j$ is $n - j$. This group is usually referred to as the group of integers modulo n.  My understanding of this is as following, please let me know if I have gotten the main idea.  Let's pick $Z_3$ as an example.  Then the members of this group is $0,1,2$ let $j=2$ then $n-j=3-2=1$  it says the inverse of $j$ is $n - j$, by definition, an inverse in a group is such ab=identity. And here the identity is 0 under addition.   therefore, 2+1=3 and 3 mod 3 =0 which is the identity.   This applies to any n in $Z_n$  So, it satisfies as a group?
81513
I want to prove that:     If $F$ is a free abelian group of rank $n$, then $\text{Aut}(F)$ is isomorphic to the multiplicative group of all $n\times n$ matrices over $\mathbb Z$ with $\text{det}=\pm1$.   What I have tried:  Since $F$ is a free abelian group of rank $n$ so I can write $F=\langle x_1,x_2,...,x_n\rangle$. If $\phi\in\text{Aut}(F)$ then it should map any $x_i$ to another element $x_j$ of the basis $X=\{x_1,x_2,...,x_n\}$. My idea is to map any element of $\text{Aut}(F)$ to a suitable matrix. For example when $n=2$: $$\phi_1:=\{x_1\to x_1, x_2\to x_2\}\Longrightarrow \left( \begin{array}{cc} 1 &amp; 0 \\ 0 &amp; 1 \end{array} \right)\\ \phi_2:=\{x_1\to x_2, x_2\to x_1\}\Longrightarrow \left( \begin{array}{cc} 0 &amp; 1 \\  1 &amp; 0 \end{array} \right)$$  Am I on a right approach? Or there is another better way? Thanks.
68822
Question: If $R$ is a finite boolean ring, then show $R \cong \mathbb Z_2 \times \mathbb Z_2 \times \cdots\times \mathbb Z_2$.   I know that   $\mathrm{char}(R) =2$  $R$ has $2^k$ elements for some $k$  (this is because, if $p\neq 2$ is a prime with $p\mid|R|$, then there exists a non-zero $x \in R$ s.t. $px=0$ and $p $ is odd $\implies x=0$, contradiction).   Does this suffice to answer my question?
23025
How many elements are in the quotient ring $\displaystyle \frac{\mathbb Z_3[x]}{\langle 2x^3+ x+1\rangle}$ ?  I guess I should be using the division algorithm but I'm stuck on how to figure it out.
89718
Show $\operatorname{Aut}(C_2 \times C_2)$ is isomorphic to $D_6$ (the group with $x^3=1$, $y^2=1$ and $xy=yx^2$).  I'm not really sure how to express the elements of $\operatorname{Aut}(C_2 \times C_2)$. Would it be sufficient to show the elements of $\operatorname{Aut}(C_2 \times C_2)$, find their order and show they bijectively map to every element of $D_6$ and satisfy $xy=yx^2$?
14261
Find the number of elements of order two in the symmetric group $S_4$ of all permutations of the four symbols {$1,2,3,4$}.    the order two elements are two cycles.number of $2$ cycles are $6$.but the given answer is $9$.where I am wrong?can anybody help me.  more generally is there any formula for the problem "Find the number of elements of order $r$ in the symmetric group $S_n$"
66569
Prove that there is no group $G$ s.t. $\operatorname{Aut}(G)=\mathbb{Q}$  I get the feeling that we should proceed by contradiction.   So let $G$ be a group s.t. $\operatorname{Aut}(G)=\mathbb{Q}$. Then we can identify elements of $\mathbb{Q}$ with automorphisms of $G$... and identities such as $\frac{1}{2}*2(g)=1(g)=g$  Can somebody help me find a contradiction?
105855
After a week in my Abstract Algebra class, the professor proposed this as a problem. I'm not entirely sure where to begin. $ V = \{ e, \tau, \tau_1, \tau_2 \}$, so I'm not sure exactly what is meant by $\text{Aut} (V)$. Is it simply saying that the only way to have an automorphism on $V$ is to rearrange the order of the elements of $V$ and therefore $\text{Aut} (V) \cong S_3$?
22367
I need help with this:  If a is an element of $\mathbb Z_n$ and $\gcd(a,n) &gt; 1$ , then a is not invertible.   First you show that if $a$ is an element of $\mathbb Z_n$ and $\gcd(a,n)&gt;1$, then there  is an element $b$ of $\mathbb Z_n$ and ($b$ is not equal to the zero element) from which $ab = 0$.   The second part I cant get. i.e. show if $b$ is not the zero element  and $ab=0$ then $a$ is not invertible.
42635
I'm interested in the number of binary matrices of a given size that are distinct with regard to row and column permutations.  If $\sim$ is the equivalence relation on $n\times m$ binary matrices such that $A \sim B$ iff one can obtain B from applying a permutation matrix to A, I'm interested in the number of $\sim$-equivalence classes over all $n\times m$ binary matrices.  I know there are $2^{nm}$ binary matrices of size $n\times m$, and $n!m!$ possible permutations, but somehow I fail to get an intuition on what this implies for the equivalence classes.
44951
How do you describe the quotient group $\mathbb{Z} \times \mathbb{Z} / \langle (1, 2) \rangle$? It is easy to get it when it is $\mathbb{Z}_n$, but it is difficult when it is $\mathbb{Z}$.  How should I start and solve the problem?