Could someone please help me correct my following proof?     Prove that if $G$ is a cyclic group of order $n$, then $G$ is isomorphic to $\mathbb{Z}_{n}$.   Proof: Let $G=\left \langle a \right \rangle$ with $|G|=n$. Define the relation $\phi:\mathbb{Z}_{n}\to G$ by $\phi(k)=a^{k}$ with $0\leq k&lt;n$.  Well-defined: For $0\leq k_{1}&lt;n$ and $0\leq k_{2}&lt;n$, if $k_{1}=k_{2}$, then $\phi(k_{1})=a^{k_{1}}=a^{k_{2}}=\phi(k_{2})$.  Operation preserving: Since $\phi(k_{1}+k_{2})=a^{k_{1}+k_{2}}=a^{k_{1}}a^{k_{2}}=\phi(k_{1})\phi(k_{2})$, the operation is preserved.  One-to-one: For $m,n\in \mathbb{Z}_{n}$, let $m&gt;n$ and assume $\phi(m)=\phi(n)$. Then $a^{m}=a^{n}$ or $a^{m-n}=e$. Then $m-n&gt;0$.     I am not sure how to prove that $m-n&lt;0$. Is it due to $G$ being abelian?   Onto: $\phi$ is onto iff for all $a^{k}\in G$, there exists a $k\in \mathbb{Z}_{n}$ such that $\phi(k)=a^{k}$. By definition, $\phi$ is onto.
123460
I want to prove that 2 cyclic groups of the same order are always isomorph. So, let $G,H$ be cyclic groups with $|G| = |H|$. Then: I suspect the isomorphism would be the one that maps a generator g to a generator h. Thus, consider this function:  $f: G \rightarrow H$ defined by $f(g^k) = h^k$. I already showed that this mapping is surjective and a homomorphism, but I get stuck at the injectivity part. Can someone give a hint (my attempts: I tried to deduce whether the kernel is $e_G$ and I also used the definition of injectivity but I got stuck in both approaches)?
100258
Let $G$ be an Abelian group of order $mn$ where $\gcd(m,n)=1$.   Assume that $G$ contains an element of $a$ of order $m$ and an element $b$ of order $n$.   Prove $G$ is cyclic with generator $ab$.    The idea is that $(ab)^k$ for $k \in  [0, \dots , mn-1]$ will make distinct elements but do not know how to argue it.   Could I say something like $&lt;a&gt;=A$, $&lt;b&gt;=B$, somehow $AB=\{ ab : a \in A , b \in B \}$ and that has order $|A||B|=mn$?  Don't know if it's the same exact or similar to .
115187
Let $G$ be a group such that for all $x\in G$ we have $x^2=e$. Show that if $G$ is finite then the order of $G$ is $2^n$.  Here is the solution I have seen in a book.      If G is finite, it can be considered as a vector space over the field $\Bbb{Z}/2\Bbb{Z}$, and then necessarily is finite dimensional, which gives $G$ as a vector space isomorphic to $(\Bbb{Z}/2\Bbb{Z})^n$.   I don't understand why $G$ can be considered as a vector space over the field $\Bbb{Z}/2\Bbb{Z}$. Can, someone, explain this? Please.
77054
I am trying to show the isomorphism  $$\mathbb Z^n/\langle (a,...,a) \rangle \cong \mathbb Z^{n-1} \oplus \mathbb Z/\langle a \rangle.$$  I've tried to define $\psi:\mathbb Z^n \to \mathbb Z^{n-1} \oplus \mathbb Z/\langle a \rangle$ an epimorphism with $\ker(\psi)=\langle (a,...,a) \rangle$ so to apply the first isomorphism theorem, but I couldn't come up with an appropiate morphism. Any hints or suggestions would be greatly appreciated.
96801
Let $G$ and $K$ be groups and $f:G\rightarrow K$ be a group   homomorphism. Suppose $g\in G$ has finite order $n$. Prove that   $\mathrm{ord}(f(g))$ is finite and divides $n$.   How can I fix this proof?  We know $g^n=e$. Since $f$ is a homomorphism, then $[f(g)]^n = f(g^n)=f(e)=e$. Let $k\mid n$ and $j\mid n$. Then we can choose the smallest $k$ such that $[f(g)^k]^j=[f(g)]^n$. Thus, $\mathrm{ord}(f(g))=k$ where $k\mid n$.  This logic seems shaky. How can I improve it? Thanks!
106228
This is related to this .  I just didn't want a prolonged discussion in the comments.  Let $\phi: G \to G'$ be a homomorphism.  Let $G$ be a finite group.  Let $K \leq G$ be the kernel of $\phi$.  Let $I \leq G'$ be the image of $\phi$.  Let $H' \leq G'$.  Find a formula relating the order of $\phi^{-1}(H')$ in terms of $H', I, K$.    Attempt at a solution: $|\phi^{-1}(H')|=|H'\cap I|\cdot |K \cap \phi^{-1}(H')|$.  My justification for this is as follows.  $\phi^{-1}(H')/(K \cap \phi^{-1}(H'))\cong H' \cap I$
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.
112845
Let $ G $ be an abelian group of order 6.  Then $ G $ has one element of order 6.  And so, $ G $ is cyclic and isomorphic to $ \mathbb Z _6 $.  In general, It is not true that an abelian group of order $n$ has an element of order $n$.  However in the case of $n=6$, it is true. I'm wondering there is any rule of the case (for example, a rule of $n$ that makes an abelian group cyclic).
120983
Prove that if $f : G \rightarrow H$ is a group isomorphism, then for any element $x \in G$ one has $o(x) = o(f(x))$, where $o(a)$ denotes the order of the element $a$.  I feel that this proof should be online somewhere but haven't been able to find anything so would be helpful if anyone could point me in the right direction.
27021
Suppose you have a group isomorphism given by the first isomorphism theorem: $$G/\ker(\phi) \simeq \operatorname{im}(\phi)$$ What can we say about the group $\ker(\phi)\times \operatorname{im}(\phi)$? In particular, when does the following hold: $$G\simeq \ker(\phi)\times \operatorname{im}(\phi)?$$ I ask this question because i want to prove that $GL_n^+(\mathbb{R}) \simeq SL_n(\mathbb{R}) \times \mathbb{R}^*_{&gt;0}$, with $GL_n^+(\mathbb{R})$ the group of matrices with positive determinant. I proved that $SL_n(\mathbb{R})$ is a normal subgroup and that $GL_n^+(\mathbb{R})/  SL_n(\mathbb{R}) \simeq \mathbb{R}^*_{&gt;0}$, using the surjective homomorphism $\det(M)$. I tried something with semidirect products but I got stuck.
59476
Let $G_1, G_2$ and $G_3$ be groups.  Let $\phi: G_1 \to G_2$ and $\sigma: G_2 \to G_3$ be isomorphisms of groups.  Show that $$\sigma\circ\phi: G_1 \to G_3$$ is an isomorphism.  I understand to prove the composition is a homomorphism (operation preserving) and a bijection.  I need help with notation and how to show the operation preserving, onto and one-to-one.  Operation preserving:  $\sigma\circ\phi(ab)=\sigma(\phi(ab))=\sigma(\phi(a)\phi(b))=(\sigma\circ\phi(a))(\sigma\circ\phi(b))$  Am I on the right track?
56373
Say a cyclic group of order $9$ is denoted by $G = \langle g \rangle$. I have a claim that  $\langle g^k \rangle = G $ whenever $k$ is a unit in $\mathbb{Z}_9$. I'm struggling to understand why this is true. Is this specific to order 9? In general, for a group of order $n$, do we require that $k$'s multiplicative inverse in $\mathbb{Z}_n$ be $n$ itself? Otherwise the generating group will be of order less than $n$, and thus cannot be equal to the group.
117977
I am doing the following exercise from [Birkhoff and MacLane, A survey of modern algebra]:     Let $G$ be a group of order $pq$ ($p,q$ primes).  Show that either $G$ is cyclic or contains an element of order $p$ (or $q$).  In the second case, show that $G$ contains either $1$ normal or $q$ conjugate subgroups of order $p$.  In the latter case the $pq-(p-1)q=q$ elements not of order $p$ form a normal subgroup.  Infer that $G$ always has a proper normal subgroup.   I think I was able to prove $G$ has a proper normal subgroup, but without quite establishing that the number of subgroups of order $p$ is 1 or $q$.  Here is my solution:    Suppose $G$ is cyclic and $G=\langle x \rangle$.  Then, $\langle x^p \rangle$ has order $q$ and is normal, so we are done.  So suppose $G$ is not cyclic.  Let $x \in G-1$.  By Lagrange's theorem, the order of $x$ is $p$, say.  Let $H:=\langle x \rangle$.  $G$ acts on all subgroups of order $p$ by conjugation.  In this action, the stabilizer of $H$ is the normalizer $N_G(H)$.  Since $N_G(H) \supseteq H$, $|N_G(H)|$ equals $p$ or $qp$.  Hence the orbit containing $H$ has length $q$ or 1 (by the orbit-stabilizer lemma).  If this value is 1, then $H$ is normal, so we are done.  If this value is $q$, then there are $pq-(p-1)q=q$ remaining nonidentity elements in $G$ that are not in any conjugate of $H$.    Case 1: If any of these $q$ elements, say some $y \in G$, has order $p$, then $K:=\langle y \rangle$ is either normal (so we are done) or has $q$ conjugate subgroups (in which case the conjugates of $H$ and of $K$ together contain $q(p-1)+q(p-1)+1&gt;qp$ elements, a contradiction).    Case 2:  If none of the $q$ elements has order $p$, then they all have order 1 or $q$, and hence form a cyclic subgroup $K$ of order $q$.  $K$ must be normal, since otherwise $g^{-1}Kg \ne K$ for some $g \in G$, whereas $K$ already exhausted all the elements of $G$ of order $q$.   Thus, $G$ has a normal subgroup of order $p$ or $q$.  While I have shown the group is not simple, I think I haven't shown yet that the number of subgroups of order $p$ is 1 or $q$.  The proof only establishes that the number of distinct conjugates of $H$ is 1 or $q$.  Any suggestions on how to prove (using elementary methods) that any other subgroup of order $p$ would have to be conjugate to $H$? In other words, I need to rule out Case 1 in the proof.
93492
My goal is to prove this:     If $G$ is a finite abelian $p$-group with a unique subgroup of size $p$,   then $G$ is cyclic.   I tried to prove this by induction on $n$, where $|G| = p^n$ but was not able to get very far with it at all (look at the edit history of this post to see the dead ends). Does anyone have any ideas for a reasonably elementary proof of this theorem?
73658
I am looking for a correct proof of this statement: If $G$ is a group such that $G/Z(G)$ is cyclic, then $G$ is commutative.  Proof: $G/Z(G)$ is isomorphic to $\operatorname{Inn}(G)$ and is cyclic, and then for every $a$ and $b$ in $G$ the inner isomorphisms $\gamma_a$ and $\gamma_b$ satisfy $\gamma_a \gamma_b = \gamma_{ab} = \gamma_{ba} = \gamma_b \gamma_a$, and therefore for every $a,b \in G$, $ab = ba$.  Is that proof complete, or am I missing something? Thanks a lot for the help.
6343
I'm asked to show that $G=GL_2(\Bbb Z/2\Bbb Z)$ is isomorphic to $S_3$. I have few ideas but I don't manage to put them all together in order to obtain a satisying answer. I first tried using Cayley's theorem ($G$ is isomorphic to a subgroup of $S_6$), and I also noticed that $\operatorname{Card}(G)=\operatorname{Card}(S3)=6$ &amp; that they're both non-abelian group.  Is this enough to say that considering $S_3$ is a subgroup of $S_6$ with the same cardinality than $G$, it has to be isomorphic to it ?  Could anyone give me some elements to get a more rigorous proof or lead me to an other path to show this statement ?  Thanks in advance
115829
I am a beginning learner of group theory.  Which of the following are isomorphic:$$\mathbb{Z_{24}}, D_{4}\times \mathbb{Z_{3}},A_{4}\times \mathbb{Z_{2}},\mathbb{Z_{2}}\times D_{6}, \mathbb{Z_{12}}\times \mathbb{Z_{2}},D_{12}, S_{4}$$  I know $\mathbb{Z_{24}}$ is cyclic and none of $\mathbb{Z_{12}}\times \mathbb{Z_{2}},D_{12}, S_{4}$ are cyclic since none has an element of order 24. I also know $D_{12}$ and $S_{4}$ are not isomorphic because $S_{4}$ has no element of order 12. Finally, I know $A_{4}$ and $D_{6}$ are also not isomorphic because $D_{6}$, unlike $A_{4}$, contains an element of order 6.  Other than this I don't know how to proceed. Any hints are appreciated. Please allow me to do the work.