My attempt : Suppose $H$ is another subgroup of $S_5$ of order $60$  by Lagrange's Theorem $\vert H\cap A_5\vert ||H|$, then $|H \cap A_5| \in\{1,2,3,5,6,10,20,30,60\}$  $$\Rightarrow |HA_5| =\frac{|H||A_5|}{|H \cap A_5|}=3600,1800,1200,900,720,600,360,180,120,60$$   So, the only candidates for the order of $|H \cap A_5|$ are $30$ and $60$.  Then I want to claim that the order must be $60$. thus $|H \cap A_5|=|A_5|$ $$\Rightarrow H=A_5$$  But I didn't know how to show the former case doesn't hold.  Please give me a hint please!  PS. I haven't learned the concepts of normal group and simple group. So, I would like you to explain without using these concepts! Thank you.
2919
How to prove that the only subgroup of the symmetric group $S_n$ of order $n!/2$ is $A_n$?  Why isn't there other possibility?  Thanks :)
52402
Let $G$ be a finite Group, let S=$\{s_1,s_2,s_3 ... |s_i\in G $ has odd order $\}$. Let $H=\langle S\rangle$ The question is:   Prove that $H$ is normal subgroup of $G$. prove that $G/H$ has order $2^k$ for $k \in \mathbb{N}$.   I tried look at $$gHg^{-1}$$ but can't understand why $gHg^{-1} \subseteq H$ probably because I cant understand what unique in H. Thanks much for help.
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.
90245
I have what seems to me a very simple proof that $A_{n}$ is the only subgroup of $S_{n}$ of index 2.  Since I've seen other people prove it with what feel like really complicated methods (), I'm wondering if I've overlooked something.  Proof: Let $H$ be any subgroup of index 2 in $S_{n}$.  Then $H\cap A_{n}$ is a normal subgroup of $A_{n}$ and since $A_{n}$ is simple, $H$ is either the trivial or the improper subgroup of $A_{n}$.  If it's the trivial subgroup it doesn't have index 2 in $S_{n}$, and otherwise the theorem is proved.
48825
I want to show that if $g,h\in G$ are group elements with finite coprime orders $m$ and $n$ and $gh=hg$ then the order of $gh$ is $mn$.    We have that  $$ (gh)^{mn}=g^{mn}h^{mn}=1 $$  using $gh=hg$, so $|gh|\leq mn$. On the other hand, $mn=\operatorname{gcd}(m,n)\operatorname{lcm}(m,n)=\operatorname{lcm}(m,n)$, since $m$ and $n$ are coprime. I am not sure how to proceed, I tried assuming that $|gh|&lt;mn$ and arriving at a contradiction, but it did not take me anywhere.
15424
Here my solution: Suppose there exists and $H \leq A_4 \oplus Z_3$ such that order of H is 18. Now, notice index of H in $A_4 \oplus Z_3$ is 2. therefore, H is normal, and therefore, the $A_4 \oplus Z_3 / H$ exists. Now, for every $\alpha$ in $A_4 \oplus Z_3$, $\alpha^2H = (\alpha H)^2 = H$ since the quotient group has order 2. But notice, $A_4$ has only 3 elements of order 2, therefore, $A_4 \oplus Z_3$ have only 3 elements of order 2, which implies H has 3 elements of order 2 and the rest must have order 1, and this is an absurd.  Is this a correct solution? Do you guys have any other solution?   thanks,
42433
I'm trying to figure out this problem:  Problem: Let $G$ be a group of order $p^k q$ with $p$ and $q$ distinct primes and $k \in \mathbb{N} \setminus \left\{0\right\}$.   a) Prove: $G$ has a subgroup of order $p^k$, or $p$ divides $|Z(G)|$. (Here $Z(G)$ means the center of $G$).   b) Prove: if $p$ divides $| Z(G)|$, there there exists an $x \in Z(G)$ of order $p$.   Attempt: For a) I tried proceeding by induction.  Base step: Assume $k=1$. Then $|G| = pq$ with $p \neq q$. Case 1: if $G$ is commutative, then $Z(G) = G$ and then $p$ divides $|Z(G)|$.   Case 2: Assume $G$ is not commutative. Then the center of $G$ is trivial, i.e. $Z(G) = \left\{e_G\right\}$ (this is a theorem of group theory). So $p$ does not divide $|Z(G)|$. Then we must show that $G$ has a subgroup of order $p$. Let $H$ be a non-trivial subgroup of $G$, that is $H \neq G$ and $H \neq \left\{e\right\}$. Then by Lagrange, $|H|$ divides $pq$. So we must have either that $|H| = p$ or $|H| = q$ (is this correct? Since $H$ is non-trivial). If $|H| = p$ then we are done. If $| H | = q$, then $|G / H| = p$ and so $G/H$ is cyclic. I want to derive a contradiction from this, and so show that $H$ cannot have order $q$, but I don't know how.  Induction step: Assume the claim holds for $|G| = p^{k-1} q$. We must then show that it also holds for $|G| = p^k q$.  I want to use the class equation. We have that $$ p^k q = |Z(G)| + \sum_{i=m+1}^n |Cl(a_i)| $$ where $|Cl(a_1)| = \ldots = |Cl(a_m)| = 1$ and $|Cl(a_i)| &gt; 1$ for $i &gt; m $. Since $p$ is a divisor of the LHS, it must also divide the linear combination. But how can I conclude from this that $p$ divides $| Z(G)|$ ? I don't know how to use the induction hypothesis.   b) Assume $p$ divides $|Z(G)|$. I first assume that $G$ is abelian. Then by structure theorem for finite abelian groups, we know that $$ G \cong \mathbb{Z}_{p_1^{k_1}} \oplus \ldots \oplus \mathbb{Z}_{p_r^{k_r}}. $$ This means that $$ p^k q = p_1^{k_1} \ldots p_r^{k_r}. $$ So $p | (p_1^{k_1} \ldots p_r^{k_r})$ and so since $p$ is prime, $p$ must divide at least one factor. Assume that $p$ divides $p_i^{k_i}$. Since $p_i^{k_i} = p_i \ldots p_i$ with $k_i$-amount of factors, we must again have that $p$ divides $p_i$. So $p = p_i$.   Now, since every $\mathbb{Z}_{p_i^{k_i}} = \mathbb{Z}_p^{k_i}$ is abelian, we have that the center equals the whole group. Let $x_i$ be an element in this center. Then $order(x_i) | p^{k_i}$ and so $order(x_i) | p$. Hence the order of $x_i$ equals $p$ in $\mathbb{Z}_p^{k_i}$. Since $x_i$ is also in $Z(G)$, this concludes the proof?   And what about the case that $G$ is not abelian?  Help with this problem is appreciated!
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.
113638
Let $G$ be a finite group. Let $H \trianglelefteq G$, with $\vert H \vert = p$, a prime, where $p$ is the smallest prime dividing $\vert G \vert$. Prove that $H \leq Z(G)$. (Hint: If $a \in H$, by normality, its conjugacy class lies inside $H$.)   My approach so far:  Let $a \in H$. Then $C(a) = \{gag^{-1} : g \in G\} \subseteq H$. Now, since $\vert H \vert$ is prime, it follows that $H$ is cyclic and moreover $H$ is abelian, hence $C_H(a) = H$ for all $a \in H$ and equivalently $[H: C_H(a)] = 1$ for all $a \in H$.  Try to show $\vert C(a) \vert = [G:C_G(a)] = [H:C_H(a)] = 1$.  So, since $\vert C(a) \vert = 1$ for all $a \in H$, we have $C(a) = \{a\}$ for all $a \in H$ and equivalently $a \in Z(G)$ for all $a \in H$, so we can say $H \subseteq Z(G)$ and moreover $H \leq Z(G)$.  My main question is, how to show $[G:C_G(a)] = [H:C_H(a)]$. But also, I haven't used the fact that $p$ is the smallest prime dividing $\vert G \vert$, so is my reasoning wrong anywhere?
3031
Let $H$ be a subgroup of a group $G$. The I know that for $a,b\in G$ we have $aH = bH$ if and only if $a^{-1}b \in H$.  My question is if it is also true that $aH = bH$ if and only if $ab^{-1} \in H$? Does it matter where the $-1$ goes?  I am guessing that this is not true since we keep putting the $-1$ on the left element, but I am also thinking that it might be true because they look like the same.  If $G$ is abelian, then it doesn't matter, so a possible counter example would have to involve a non-abelian group. If $H$ is a normal subgroup, then it should also be true because then $a^{-1}bH = Ha^{-1}b$.
50837
I am beginning with Abstract Algebra and I'm trying to understand Lagrange's Theorem. The theorem reads      For any finite group $G$, the order of every subgroup $H$ of $G$ should divide the order of $G$.   It seems simple and I've used it to solve some exercices but I believe I'm missing the essence of it. Is there an example that can help me understand it better, or visualize it? Is there a geometric interpretation?
5791
I'm looking for verification:  My claim: If $G$ is a finite group and $H$ is a (proper)subgroup of index $k&gt;1$, where $k! \leq |G|$, then $G$ is not simple.  Proof: Consider the set of left cosets of $G$ by $H$:  $$G/H = \{H,g_1H,\dots,g_kH\}$$  Then $\phi: G \rightarrow S_{G/H}$ is a homomorphism where we define:  $$\phi(g)[g_iH] = g g_iH$$  ($S_{G/H}$ is the group of permutations on elements of $G/H$ and of course $S_{G/H} \cong S_k$.)  Also $\ker \phi \neq G$ because $\phi(g_1)$ is clearly not the identity in $S_{G/H}$.  Because $\phi$ is a homomorphism,  $$ G/\ker\phi \cong \text{im}\phi$$  Then $|\text{im} \phi| \leq |S_k| = k!$ and if $k! \leq |G|$ then $|\ker\phi| &gt; 1$ and $\ker\phi$ is a non-trivial normal subgroup of $G$.  Example: If a group of order $36$ has a subgroup of index $3$ or $4$ then the group is not simple.
35777
Given the following groups, what is the maximum possible order for an element for   (a) $S_5$ (b) $S_6$ (c) $S_7$ (d) $S_{10}$ (e) $S_{15}$  My book justifies the answer as          (a) The greatest order is $6$ and comes from a product of disjoint cycles of length 2 and 3          (b) The greatest order is $6$ and comes from a cycle of length $6$      The other answers were justified exactly the same way, that is (c) 12, (d) 30, (e) 105  I do not understand how in (a) we even got the number "6" from $S_5$ and what disjoint cycles they are referring to. Could someone at least justify one for me?
84099
Can somebody explain me how is the converse of the  Lagrange's theorem  false by saying that $A_4$ (alternating group) doesn't contain any subgroup of order $6$?
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.
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).
54778
Reference:   Let $G$ be a group. Let $H,K$ be subgroups of $G$. How do I prove that $[G:H\cap K]\leq [G:H][G:K]$?  Let's not assume any index is finite. Then, still the result holds? If so how do I prove it?