Let's say $\mathbb{Z}_n$ is the cyclic group. and I know that $Aut(\mathbb{Z}_n)\cong\mathbb{Z}_n^{\times}$, the multiplicative group of $\mathbb{Z}_n$. so the question is, is there a way to express $\mathbb{Z}_n^{\times}$ as a direct sum of cyclic groups?  I just thought that since the order of $\mathbb{Z}_n^{\times}$ is $\phi(n)$, and since it is abelian, maybe I should use the divisors of $\phi(n)=d_1^{s_1}...d_k^{s_k}$ like $\mathbb{Z}_{d_1^{s_1}}\oplus...\oplus\mathbb{Z}_{d_k^{s_k}}$, but I realized that there are too many non-isomorphic abelian groups with same order: for example, $\mathbb{Z}_4$ and $\mathbb{Z}_2\oplus\mathbb{Z}_2$ are both abelian and of order $4$ but they are not isomorphic.  so, yes, I guess there is a way since it is abelian, but I don't know how to do it specifically. can you give me a hint?
12906
When can we say a multiplicative group of integers modulo $n$, i.e., $U_n$ is  cyclic?   $$U_n=\{a \in\mathbb  Z_n \mid \gcd(a,n)=1 \}$$  I searched the internet but did not get a clear idea.
63390
Every cyclic group with order > 2 has at least 2 distinct generators   Here's what I've got so far:  Either order of our group, $G$, is finite or infinite.  Suppose infinite: then our group is isomorphic to $\mathbb Z$ under addition. This group has two distinct generators, therefore so does $G$.  Suppose finite: Then $G$ is isomorphic to $\mathbb Z_n$ under addition modulo $n$.  We know order of $G$ is $\geqslant 3$. If order of $G$ is odd, both 1 and 2 are generators of $\mathbb Z_n$ under addition modulo $n$, so $G$ has at least two generators.  If order of $G$ is even, then both 1 and some odd number between 1 and $n$ are generators. How do we determine this odd number?  Does this look like I'm on the right track?
112862
In the case of abelian groups, I have been treating these two set operations as more or less indistinguishable. In early mathematics courses, one normally defines $A^n := A\times A\times\ldots\times A$; however in, for example, the fundamental theorem of finitely generated abelian groups, we normally write that every such group is isomorphic to one of the form $$ \mathbb{Z}^n \oplus \mathbb{Z}_{r_1} \oplus \cdots \oplus \mathbb{Z}_{r_t}  $$ where $\mathbb{Z}^n$ now means $\mathbb{Z}\oplus\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}$.  From an intuition perspective, and in the sense of sets, is this more or less the same as $\mathbb{Z}\times\mathbb{Z}\times\cdots\times\mathbb{Z}$? (Bear in mind I am normally using these ideas in relation to homology groups.)
10044
How can I compute $\mathrm{Aut}(\mathbb{Z}/4\mathbb{Z}\times\mathbb{Z}/6\mathbb{Z})$?  If $\varphi$ is one of this automorphism then $\varphi((1,0))=(1,0),(1,3),(3,3),(3,0)$ and $\varphi((0,1))=(0,1),(0,5),(2,1),(2,5)$.  So it seems to me that this groups contains 16 elements, am i right? And if I'm right there is a way to understand to which group of cardinality 16 this is isomorphic to?  EDIT: $\mathbb{Z}/4\mathbb{Z}\times\mathbb{Z}/6\mathbb{Z}\cong\mathbb{Z}/8\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}$. And $\mathbb{Z}/8\mathbb{Z}$ and $\mathbb{Z}/3\mathbb{Z}$ are characteristic (am I right?), so $\mathrm{Aut}(\mathbb{Z}/8\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z})\cong\mathrm{Aut}(\mathbb{Z}/8\mathbb{Z})\times\mathrm{Aut}(\mathbb{Z}/3\mathbb{Z})\cong\mathbb{Z}/8\mathbb{Z}^*\times\mathbb{Z}/3\mathbb{Z}^*$ and this group has cardinality 8, right? Where is my mistake?
15252
Let $G$ be a finite abelian group of order $n$ and let $H$ be a subgroup of $G$ of order $m$. Show that $G$ has a subgroup isomorphic to $G/H$.   Here are my thoughts:  Define $\mu_n := \{z \in \mathbb{C}|z^n=1\}$ and $\mu:=\bigcup_{n=1}^{\infty}\mu_n$.  Since a finite abelian group is self-dual, we have $G/H \cong \operatorname{Hom}(G/H,\mu)$.   By the structure theorem for finite abelian groups, we have $$G \cong \mathbb{Z}_{n_1}\times \cdot \cdot \cdot \times \mathbb{Z}_{n_s}$$ where $n_{i+1}|n_i$ for $1 \leq i \leq s-1$ and $n_1\cdot \cdot \cdot n_s=n$, and $n_i \geq 2$ for all $i$.  Also, we have $$H \cong \mathbb{Z}_{m_1}\times \cdot \cdot \cdot \times \mathbb{Z}_{m_l}$$ where $m_{i+1}|m_i$ for $1 \leq i\leq l-1$ and $m_1 \cdot \cdot \cdot m_l=m$, and $m_i \geq 2$ for all $i$.   I want to somehow write $G/H$ as a product of cyclic subgroups of the $\mathbb{Z}_{n_i}$, for now denote this product by $A_1\times \cdot \cdot \cdot \times A_s$, where $A_i \leq \mathbb{Z}_{n_i}$, so that then I can write $$\operatorname{Hom}(G/H,\mu) \cong \operatorname{Hom}(A_1, \mu)\times \cdot \cdot \cdot \times \operatorname{Hom}(A_s, \mu).$$  Then, since $\operatorname{Hom}(A_i,\mu) \leq \operatorname{Hom}(\mathbb{Z}_{n_i},\mu)$ and (by an already known result) $\operatorname{Hom}(\mathbb{Z}_{n_i},\mu) \cong \mathbb{Z}_{n_i}$, I will have $$\operatorname{Hom}(G/H, \mu) \leq \mathbb{Z}_{n_1}\times \cdot \cdot \cdot \times \mathbb{Z}_{n_s}. $$  My problem is I cannot see how to write $G/H$ as such a product or if it's even possible. I appreciate any guidance and suggestions. Thanks.
52278
Let $U_n$ denote the group of units in $\mathbb{Z}_n$ with multiplication modulo $n$. It is easy to show that this is a group. My question is how to characterize the $n$ for which it is cyclic. Since the multiplicative group of a finite field is cyclic so for all $n$ prime, it is cyclic. However I believe that for certain composite $n$ it is also cyclic.   Searching through past posts turned up , where there was an answer containing the sentence "In number-theoretic situations there are coherent things that can be said, and/but in general I think nothing decisive can be said."  What are those number theoretic situations?  Thanks
22984
Let $G$ be a finite abelian $p$-group. It is quite elementary to see that if $g \in G$ is an element of maximal order (and thus its span is a cyclic subgroup of $G$ of maximal order) then $G$ can be written as the direct sum $G=\langle g \rangle \oplus H$ for some $H \leq G$ (subgroup of $G$). For a proof see  for example (page 2).  My question: Do we need that $G$ is a $p$-group or does it also work for arbitrary finite abelian groups?   I think it is wrong for general groups because I looked around quite a bit and always only found the above theorem, but I could not find a counter-example.
91972
If $\gcd(m,n)=1$, then $\mathbb{Z}_n \times \mathbb{Z}_m$ is a cyclic group.    Let's denote $\mathbb{Z}_n=\langle1_n \rangle$ and $\mathbb{Z}_m=\langle1_m \rangle.$ My proof goes as follows: since $|\mathbb{Z}_n \times \mathbb{Z}_m|=mn$ and $\gcd(m,n)=1$, $|\langle 1_n,1_m\rangle|=\text{lcm}(|\langle 1_n\rangle|,|\langle 1_m\rangle|)=mn$. Hence the direct product is cyclic. Is my proof correct?
74387
I'm trying an old problem here:   Suppose $n\mid m$, and I have a natural ring homomorphism $\varphi\colon \mathbb{Z}/m\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z}$ defined by $\varphi(j)=j\mod{n}$. I can verify that this is a ring homomorphism, but why does it induce a surjective group homomorphism on the unit groups $(\mathbb{Z}/m\mathbb{Z})^\times\to(\mathbb{Z}/n\mathbb{Z})^\times$?  My question in particular is why is it surjective? I take $k\in(\mathbb{Z}/n\mathbb{Z})^\times$. Then there exist integers $s,t$ such that $sn+tk=1$. Since $n\mid m$, I can also write $m=na$. So multiplying through I get $sm+tka=a$. I'm lost here. How can I find a unit in $(\mathbb{Z}/m\mathbb{Z})^\times$ which maps to $k$ to show surjectivity?
20701
From  I see that $\mathbb{Z}^{\mathbb{N}}$ is not a free abelian group. Someone can explain me why?  Thanks a lot!
6061
Q: Show that $(\mathbb{Z}/n)^\times$ cannot be cyclic (i.e., there is no   primitive root modulo $n$) unless $n$ is either a prime power or   twice a prime power.  I'm reading a solution. The first step is that for any $n$ it can be factorized into three relatively prime terms: $n = 2^r \cdot p^s \cdot m$ where $p$ is prime.  Since these terms are relatively prime, we can say that:  $(\mathbb{Z}/n)^\times \cong   (\mathbb{Z}/2^r)^\times \times (\mathbb{Z}/p^s)^\times \times (\mathbb{Z}/m)^\times$  The next step claims that for $(\mathbb{Z}/n)^\times$ to be cyclic, at most one of the three factored groups can be an even cyclic group. Why is this statement true? I know what cyclic groups are, does "even cyclic" just mean a cyclic group of even order?
42453
1) Is $\mathbb{Z}/\langle n\rangle$ cyclic and of order $n$? Why or why not? ($\mathbb{Z}/\langle n\rangle$ is defined to be the factor group of $\mathbb{Z}$ determined by $\langle n\rangle$.)  I guess I'm having trouble understanding factor groups (quotient groups make much more sense) and cyclic groups. My textbook isn't the best at explaining what's going on. I just chose some past homework problems that looked related to my problems.  This is my first abstract algebra course and I'm floundering a lot.
18496
In trying to understand why there are two sets of groups of order 4.  I know that there exists the Vierergruppe of $\mathbb{Z}_2 \times \mathbb{Z}_2$ and the group $\mathbb{Z}_4$ but I do not understand why they are not isomorphic
9284
I have to show that the ring $\mathbb{Z}[X]/(X^2-1)$ is not isomorphic with $\mathbb{Z}\times \mathbb{Z}$.    I know that $(\mathbb{Z}\times\mathbb{Z})^*=\{(\pm1,\pm1)\}$, so I thought I should be looking for elements which have inverses in $\mathbb{Z}[X]/(X^2-1)$ and hopefully find more or less than 4. But I didnt succeed, so I need hints. Thanks.
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?
23895
I'm currently working on a proof, but didn't get the right idea, a hint would be appreciated:  Show that $\mathbb{Z}/n^2\mathbb{Z}$ is not isomorphic to $\mathbb{Z}/n \times \mathbb{Z}/n$  So, lets assume there is an isomorphism between these groups, my first idea to was to construct a contradiction with the homomorphism property. But I don't know how, there should be an element in  $\mathbb{Z}/n \times \mathbb{Z}/n$ which is not in $\mathbb{Z}/n^2$ to reach this kind of contradiction. Of course there is $n \in \mathbb{Z}/n^2$ which is not in the other group but I don't know if this helps.  Thanks
37073
Let $G$ be a group where every non-identity element has order 2.     If |G| is finite then $G$ is isomorphic to the direct product $\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \ldots  \times \mathbb{Z}_{2}$.  Is the analogous result   $G= \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \ldots $.  true for the case |G| is infinite?