Using the fact that if $a$ and $b$ are two distinct elements of order $5$, either $\langle a \rangle \cap \langle b \rangle =\{e\}$ or $\langle a \rangle = \langle b \rangle$, show that $G$ must have an element of order $3$. Do not assume that $G$ is a cyclic group.
123260
Let G be a cyclic group of order 15. Why must G contain at least 2 elements of order 3?  Is it to do with Cauchy's theorem? Then once we know that it has one, the inverse of this element has order 3 as well?
35077
Let $G$ be a finite abelian group, and let $K$ be a subgroup of $G$. Does $G$ necessarily have a subgroup $H$ such that $H\cong G/K$ and $H\cap K=\langle 0\rangle$?  I'm not sure where to start.
64228
Let $G$ be a group and $a,b \in G$ with $a \ne 1$ and $b \ne 1$. Assuming $ab^2 = b^3a$ and $a^2=1$ I need to prove that the order of $b$ is $5$.   I have proved by contradiction that it can't be 2 or 3 but I don't know how to prove that it must be 5 and it can't be 4.
64440
I am reading the proposition on page 102 of Dummit and Foote's Algebra, 3rd edition. Could someone explain why $p$ divides $|G/N|$ on the highlighted line?     Proposition 21.   If $G$ is a finite abelian group and $p$ is a prime dividing $|G|$, then $G$ contains an element of order $p$.      Proof:   The proof proceeds by induction on $|G|$, namely, we assume the result is valid for every group whose order is strictly smaller than the order of $G$ and then prove the result valid for $G$ (this is sometimes referred to as complete induction).   Since $|G| &gt; 1$, there is an element $x \in G$ with $x \neq 1$.   If $|G| = p$ then $x$ has order $p$ by Lagrange’s Theorem and we are done.   We may therefore assume $|G| &gt; p$.      Suppose $p$ divides $|x|$ and write $|x| = pn$.   By Proposition 2.5(3), $|x^n| = p$, and again we have an element of order $p$.   We may therefore assume $p$ does not divides $|x|$.      Let $N = \langle x \rangle$.   Since $G$ is abelian, $N \trianglelefteq G$.   By Lagrange’s Theorem, $|G/N| = |G|/|N|$ and since $N \neq 1$, $|G/N| &lt; |G|$.   Since $p$ does not divide $|N|$, we must have $p \mid |G/N|$.   We can now apply the induction assumption to the smaller group $G/N$ to conclude it contains an element $\bar{y} = yN$, of order $p$.   Since $y \notin N$ ($\bar{y} \neq \bar{1}$) but $y^p \in N$ ($\bar{y}^p = \bar{1}$) we must have $\langle y^p \rangle \neq \langle y \rangle$, that is, $|y^p| &lt; |y|$.   Proposition 2.5(2) implies $p \mid |y|$.   We are now in the situation described in the preceding paragraph, so that argument again produces an element of order $p$.   The induction is complete.      (Original image .)
54985
I've figured out that if I know $G$ is not cyclic, then it for any $a \in G, o(a) \neq 4$ (or the order of any element in group $G$ is not 4).  I know ahead of time that the elements in the group ($\forall x \in G$) must have order $o(x)=k$ where $0 &lt; k \leq 4$ where $k \in \mathbb{Z}$, so $k = 1, 2, 3, 4$.  If $G$ not cyclic, then we know $k \neq 4$ so we have $k = 1, 2, 3$ left.  I know how to show that $k \neq 3$ but I am not sure if I am supposed to do it for all possible $k$ because obviously if our order was higher (let's say $n$) then it would get messy. It ends up being in this case that $k = 2$ is good and $k = 1$ is trivial. I have looked up online and it says using Lagrange's Thm we know it has to be $k = 1, 2$ since $3 \nmid 4$, but I cannot use that theorem as we have not have learned it in class. How can I show that $k = 1$ or $2$ in another way?  However working from using the result that if we have $x \in G$ and $H = \langle a \rangle$ and knowing that $\left\vert{H}\right\vert=1$ or $2$, then $x^2=1$ for either case. Then I am done right?  What is another, cleaner, better way of trying to answer the question in the title?  I apologize if my formatting is poor since I am new at LaTeX and extraordinarily bad at algebra it seems. Thank you in advance.
100106
Let $G$ be a cyclic group of finite order $n$. Then, the order of the cyclic subgroup $\langle g^k\rangle$ is $\frac {n}{\operatorname{gcd}(n,k)}$  Proof:  $G$ cyclic $\Rightarrow \exists g \in G: G = \langle g\rangle$ $\Rightarrow \operatorname{ord}(g) = \lvert G\rvert = n $  We know, because every subgroup of a cyclic group is cyclic, that $\lvert\langle g^k\rangle\rvert =  \operatorname{ord}(g^k)$. Therefore, we only need to find the order of $g^k$.  Let $l$ be the order of $g^k$. Then:  $(g^k)^l = e \iff g^{kl} = e \iff n|(kl)$  I'm now trying to show that $l = n/{\operatorname{gcd}(n,k)}$, but this seems difficult for me. I tried to assume that there was an element $l' &lt; n/{\operatorname{gcd}(n,k)}$ that satisfied the conditions, in the hope of reaching a contradiction, but this did not work out. Can anyone help?
43262
Let $G$ be a cyclic group of order $m$ generated by an element $a$.  I want to show that the order of $a^k$ is $m/d$, where $d:=\gcd(k,m)$.  I have a simple proof, but want to make sure I haven't overlooked anything since the other proofs I've seen (such as in Dummit and Foote) look more complicated.    The order of $a^k$ is the cardinality of the set $\{a^{ks}: s \in \mathbb{Z}\}$, which equals the cardinality of the set $\{a^{ks} a^{mt}: s,t \in \mathbb{Z} \}$ since $a^m$ is the identity.  But the set $\{ks+mt: s,t \in \mathbb{Z} \}$ is the set $d \mathbb{Z}$ by the basic properties of integers. So we have that the cardinality of the set is $| \langle a^k \rangle| = |\{1, a^d, a^{2d}, \ldots, a^{m-d} \}| = m/d$.
21493
Show, without Lagrange theorem, that every group $G$ of order $6$ has an element of order $3$.   I know that if $G$ has an element $a$ of order 6, then $a^2$ has order $3$, but I am not shure how to show that for the case where $G$ does not have an element of order $6$. I also know that we can only have elements $1,2,3,$ or $6$ (we proved this independent without Lagrange theorem).
16646
Suppose $a$ and $b$ are elements of a group $G$. If $a^{-1}b^{2}a=b^{3}$ and $b^{-1}a^{2}b=a^{3}$, prove $a=e=b$.   I've been trying to prove but still inconclusive. Please prove to me.  Thanks very much for proof.
76725
Say you have two groups $G = \langle g \rangle$ with order $n$ and $H = \langle h \rangle$ with order $m$. Then the product $G \times H$ is a cyclic group if and only if $\gcd(n,m)=1$.  I can't seem to figure out how to start proving this. I have tried with some examples, where I pick $(g,h)$ as a candidate generator of $G \times H$. I see that what we want is for the cycles of $g$ and $h$, as we take powers of $(g,h)$, to interleave such that we do not get $(1,1)$ until the $(mn)$-th power. However, I am having a hard time formalizing this and relating it to the greatest common divisor.  Any hints are much appreciated!
43855
Let $G$ be abelian, $H$ and $K$ subgroups of orders $n$, $m$. Then G   has subgroup of order $\operatorname{lcm}(n,m)$.   This is a statement that my lecturer mentioned in my (beginners') Abstract Algebra class. I'm not sure I understand why it's true.  What I have so far: Use the abelian group structure theorem on $\langle H, K\rangle$ (finite group generated by $H$ and $K$). Then $\langle H,K\rangle=C_{a_1}C_{a_2}\dotsm$ and $n, m |\langle H,K\rangle$. Which means it's also true that $\operatorname{lcm}(n,m)|\langle H,K\rangle$. Can I leverage this to say there's a subgroup of order $\operatorname{lcm}(n,m)$?
11401
Suppose $G$ is a finite group and I know for every $k \leq |G|$ that exactly $n_k$ elements in $G$ have order $k$. Do I know what the group is? Is there a counterexample where two groups $G$ and $H$ have the same number of elements for each order, but $G$ is not isomorphic to $H$? I suspect that there is, but I haven't thought of one.
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.
20677
In , people discussed that there cannot be cyclics subgroups of order 9 in $S_6$.   How does one reason that the subgroups isomorphic to $C_3 \times C_3$ (the other candidates to be subgroups of order 9) are precisely the ones given by two disjoint permutations $\langle (123),(456) \rangle$
123501
Consider the symmetric group $S_5$. I would like to find how many elements of $S_5$ are of order 5, and how many are of order 6. I would also like to determine what the maximum order of an element in this group would be.  Here is what I have so far: elements of order 5 are the 5-cycles. Elements of order 6 (since a 6-cycle is impossible for 5 elements) must have at least an even cycle and a cycle of length divisible by 3, and 2 + 3 = 5, so elements of order 5 are a combination of 2-cycles and 3-cycles.  How can I find the total count of such elements of order 5 and order 6, and the maximum order? I'm not entirely sure where to go from here.
58605
Question is:     Let $G$ be a group of  order 2n. Suppose half of the element of G are of order 2 and the other half form  a subgroup $H$ of order n . Prove that $H$ is of odd order  and is an abelian subgroup of $G$   What could i see is..  if we prove that order of every element of $H$ is odd , then order of $H$ is odd . Also i am unable to use the fact that half of the element of $G$ are of order 2.  Please help me to clear this.  Thank You.
79777
Let $H$ and $K$ be subgroups of $G$. Let $H$ have order $m$ and $K$ have order $n$, where $m$ and $n$ are   relatively prime. Then  $H \cap K=\{e\}$   My proof:  Let $H$ and $K$ be subgroups where the $ord(H)=m$ and $ord(K)=n$ and $m,n$ are relatively prime. We know that $H \cap K$ is a subgroup of $H$ and $K$ since it contains the elements of both in $H$ and $K$. We shall let the ord($H \cap K$)=d.   Then, by Lagrange's theorem, the order of $H$ is a multiple of the order of $H \cap K$. In other words, $m$ is a multiple of $d$ or $d|m$. Similarly,  by Lagrange's theorem, the order of $K$ is a multiple of the order of $H \cap K$. In other words, $n$ is a multiple of $d$ or $d|n$. Since $d$ divides both $m$ and $n$ and we know $m$ and $n$ are relatively prime then, the order of $H \cap K$ must be $1$.   Since $H \cap K$ is a subgroup, then by properties of subgroup, it contains an inverse which means it must also contain an identity and since $H \cap K$ contains one element, it must contain the identity. As a result, $H \cap K=\{e\}$  Hopefully, someone can confirm or correct any mistakes I made. Thanks!