I am considering the normal subgroup $H=\langle[28]\rangle$ of the additive group $\mathbb{Z}_{220}$. I want to obtain the elements of the quotient group $\mathbb{Z_{220}}/H$. Since $H=\langle14[2]\rangle$, we have that $|H|=\frac{220}{\gcd(220,14)}=55$. Therefore, by Lagrange's theorem, $|\mathbb{Z}_{220}/H|=\frac{|\mathbb{Z}_{220}|}{|H|}=\frac{220}{55}=4$. So, I know that $\mathbb{Z}_{220}/H$ has four elements and one of them is $[0]+H$. The problem I encounter is that, for example, $[0]+H$ is the same as $[28]+H$. Because of this, I have to be careful when writing the other three elements to make sure that I am not making the mistake of writing twice the same element because it is written in two ways that &quot;look different&quot;. If I wrote explicitly all $55$ elements of each element (maybe I should call it class or coset) of the quotient group then I would be sure I am not making that mistake. But this does not look very efficient. I was wondering if there might be some clever way of choosing the representative of each class so as to make sure there are no hidden repetitions. Thanks in advance.
23106
Suppose that $H$ is a (cyclic) subgroup of order $m$ of a cyclic group   $G$ of order $n$. What is $G/H$?   It's a very simple question but I am still struggling with getting accustomed to the notion of the quotient group.   Now, I know that $m$ divides $n$ and that $G/H$ will be a group of cosets such that $\{gH:g\in G\}$. As a result, there will be $n/m$ of such cosets. I was wondering I can say anything more specific than just that?
65821
Problem: Prove there are no simple groups of order $1,000,000$.   So, I have used Sylow's Theorem to get that the number of Sylow $5$-subgroups is either $1$ (and we're done) or $16$.  I am assuming we have $16$, and let $H$ and $K$ be two of these Sylow $5$-subgroups. I'm trying to show $H \cap K$ is normal in $G$.  So, $|H\cap K| = |H||K|/|HK| &gt; |H||K|/|G| \approx 244.14$  So since $H\cap K&lt;H$, $|H\cap K|$ divides $|H|$. Thus, $|H \cap K| = 625$ or $3125$.  IF $|H\cap K| = 3125$, then $[H \cap K:H] = 5$ which is the smallest prime divisor of $15625$. So $H \cap K$ is normal in $H$ and normal in $K$. This forces $HK$ to be contained in the normalizer of $H \cap K$. Since the normalizer of $H \cap K$ is a subgroup of $G$, then its order must divide $G$. The size of $HK$ is $3125^2$, which is contained in the normalizer. So $3125^2$ must divide $1,000,000$. Contradiction.   So the normalizer of $H \cap K$ is all of $G$, and $H \cap K$ is normal.  I am completely stuck on what to do if $H \cap K$ has order $625$. Any help would be much appreciated.
96326
Well, first, let $x$ be an element of the factor group $\mathbb Q/\mathbb Z$ ($\mathbb Q$ and $\mathbb Z$ are the additive groups, by the way).  Now, how do we find an integer n, such that, $nx = 0$, $n \ge 0$?  I managed to figure out that each x can be uniquely represented as $x= \dfrac{a}{b} + \mathbb Z$ for $0 \le a&lt; b &lt; 1 , \; \text{gcd}(a,b)=1$.  Now, I've considered that perhaps, I'm not understanding the question. When $nx = 0$, I assume it means, essentially the same thing as $nx =\mathbb Z$, which is the identity element of Q/Z, right?  The obvious choice for $n$ would be a multiple of $b$.  Trying $n=b$, we have, $nx = a + b \mathbb Z $, but this is just a single coset of $b \mathbb Z$. A particular equivalence class $(a \mod b)$.  Which brings me to another question:  Consider $x+x = \dfrac{a}{b}+ \mathbb Z + \dfrac{a}{b} + \mathbb Z = 2 \dfrac{a}{b}+ \mathbb Z $,  But, $2x = \frac{2a}{b} + 2 \mathbb Z $.  I know, I must have slipped with the permissible algebra, but somehow, I just seem to be unable to reconcile the $2$, or choose between them.  Which of these is correct? (Perhaps the first one, given that addition is actually the operation of $\mathbb Q/\mathbb Z$?)  Also, how would I go about finding the integer $n$? I would really appreciate it if, instead of actually finding $n$, someone could hand me a pointer in the right direction, and let me wrangle it myself.  Thank You!
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.
12270
I have a group $G$ with $p^2$ elements, where $p$ is a prime number. Some (potentially) useful preliminary information I have is that there are exactly $p+1$ subgroups with $p$ elements, and with that I was able to show $G$ has a normal subgroup $N$ with $p$ elements.   My problem is showing that $G$ is abelian, and I would be glad if someone could show me how.   I had two potential approaches in mind and I would prefer if one of these were used (especially the second one).   First: The center $Z(G)$ is a normal subgroup of $G$ so by Langrange's theorem, if $Z(G)$ has anything other than the identity, it's size is either $p$ or $p^2$. If $p^2$ then $Z(G)=G$ and we are done. If $Z(G)=p$ then the quotient group of $G$ factored out by $Z(G)$ has $p$ elements, so it is cylic and I can prove from there that this implies $G$ is abelian. So can we show theres something other than the identity in the center of $G$?  Second: I list out the elements of some other subgroup $H$ with $p$ elements such that the intersection of $H$ and $N$ is only the identity (if any more, due to prime order the intersected elements would generate the entire subgroups). Let $N$ be generated by $a$ and $H$ be generated by $b$. We can show $NK= G$, i.e every element in G can be wrriten like $a^k b^l $. So for this method, we just need to show $ab=ba$ (remember, these are not general elements in the set, but the generators of $N$ and $H$).   Do any of these methods seem viable? I understand one can give very strong theorems using Sylow theorems and related facts, but I am looking for an elementary solution (no Sylow theorems, facts about p-groups, centrailzers) but definitions of centres and normalizers is fine.
3691
Please excuse the selfishness of the following question:     Let $G$ be a group and $H \le G$ such that $|G:H|=2$. Show that $H$ is normal.   Proof:  Because $|G:H|=2$, $G = H \cup aH$ for some $a \in G \setminus H$.   Let $x\in G$. Then $x \in H$ or $x \in aH$.  Suppose $x \in H$. Then $xhx^{-1} \in H$ because $H$ is a group.  Suppose $x \in aH$. Then $ahH(ah)^{-1} = ahHh^{-1}a^{-1} = a(hHh^{-1})a^{-1}$ ***1  ***1: Can I say now that $x \in H$, based on the that $hHh^{-1} \in H$ and that $aa^{-1} = e$ ?  Thank you for the time you spent reading my doubts.
31296
Evidently, $(\mathbb{Z} \oplus \mathbb{Z})/ \langle (2,2) \rangle$ has order $4$, but I think it's infinite.   The four cosets are listed as $(0,0) +  \langle (2,2) \rangle$, $(0,1) +  \langle (2,2) \rangle$, $(1,0)+  \langle (2,2) \rangle$ and $(1,1) +  \langle (2,2) \rangle$. However, $(2,0)$ doesn't appear to be in any of these cosets. Maybe the answer I'm being told is wrong.
80770
Just wanted some feedback to ensure I did not make any mistakes with this proof. Thanks!  Since $G$ is abelian, every subgroup is normal. Since $G$ is simple, the only subgroups of $G$ are $1$ and $G$, and $|G| &gt; 1$, so for some $x\in G$ we have $x\neq 1$ and $\langle x\rangle\leq G$, so $\langle x\rangle = G$. Suppose $x$ has either infinite order or even order. Then $\langle x^2\rangle \leq G$ but $\langle x^2\rangle \neq \langle x \rangle$, a contradiction. So $x$, and therefore $G$, has odd order, and by Feit-Thompson, $G\cong Z_p$ for some prime $p$.  Edit: Thanks, I see that Feit-Thompson is too much.  Since $G$ is abelian, every subgroup is normal. Since $G$ is simple, the only subgroups of $G$ are $1$ and $G$, and $|G| &gt; 1$, so for some $x\in G$ we have $x\neq 1$ and $\langle x\rangle\leq G$, so $\langle x\rangle = G$. Suppose $x$ has infinite order. Then $\langle x^2\rangle \leq G$ but $\langle x^2\rangle \neq \langle x \rangle$, a contradiction. So $x$, and therefore $G$, has finite order. Suppose $x$ has composite order $n$ so for some $p &gt; 1$ that divides $n$, $\langle x^p \rangle$ is a proper non-trivial subgroup of $G$, so $G$ is not simple. So $G$ is a cyclic group of prime order.
44325
I'm trying to show the following     $\langle (1,2)(3,4), (1,2,3,4,5) \rangle = A_5$   I managed to prove this but I think my solution is very inelegant. Here's my argument  let $J = \langle (1,2)(3,4), (1,2,3,4,5) \rangle,$ then $\lvert J \rvert \geq 8$ and $\lvert J \rvert$ divides $\lvert A_5 \rvert = 60,$ so the possibilities are  $\lvert J \rvert = 10, 12, 15, 20, 30, 60$.  Then I sat down and calculated 13 more elements of $J$, so now the possibilities are $\lvert J \rvert = 30$ or $60$. But we can't have $\lvert J \rvert = 30$ because then $J$ would be normal (index 2 theorem) which would contradict $A_5$ being simple, so we must have $J = A_5$.  The calculation part make this proof quite long winded, is there a simpler way of getting the result?
40824
Those 5 Statements are all equivalent to the Statement that $H$ is a normal subgroup of $G$:  $(1) \forall g \in G, h\in H$ we have $ghg^{-1} \in H$  $(2) \forall g \in G, gHg^{-1} \subseteq H$  $(3) \forall g \in G, gH = Hg$  $(4)$ Every right coset of $H$ is a left coset  $(5) H$ is the kernel of a homomorphism of $G$ to some other Group  In class we used the Definition 3 can someone explain me why 3 and 1 are equivalent?  Edit: It would be convenient to show the equivalence of all 5 statements
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!
65300
Let $SL_2(\mathbb{Z})$ denote the group (under usual matrix multiplication) of $2\times2$ matrices with integer entries and determinant $1$. Let $H$ be the subgroup of $SL_2(\mathbb{Z})$ consisting of those matrices such that the diagonal entries are all equivalent to $1 \pmod 3$ and the off-diagonal entries are all divisible by $3$.   What is the index of $H$ in $SL_2(\mathbb{Z})$? There are a total of $3^4=81$ different equivalence classes of matrices in $SL_2(\mathbb{Z})$ modulo $3$ (each of the entries can have $0,1,2$ as remainders). Now, the given condition implies only one of the possible $81$ combinations modulo $3$. How do we proceed? any hints? Thanks beforehand.
63504
I'm trying to describe the groups with presentations $\langle x,y\ |\ xy=yx,x^5=y^3\rangle$ and $\langle x,y\ |\ xy=yx,x^4=y^2\rangle$. I have some problems getting a good picture of what they look like...   For the first one, $xy=yx$ allows us to say that any element in the group can be written as $x^iy^j$ for some $i,j$. More precisely, using $x^5=y^3$, we can always reduce $j$ to $0,1$ or $2$, whereas $i$ could have any value. So I was thinking about $\mathbb{Z} \times C_3$. But the other way around (reducing $i$ this time) it's also $C_5 \times \mathbb{Z}$. But then I should be able to prove that both descriptions are equivalent, which I can't do...  For the second, I proceed similarly and get to the same kind of problem...  Could you tell me where I'm going wrong?  Thank you very much in advance!
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?
28394
Consider the quotient $\mathbb{Z}-\text{module}$ $$M=\frac{\mathbb{Z}^3}{\langle (3,3,1),(2,2,2)\rangle}$$ Prove that $M$ is isomorphic to $\mathbb{Z} \oplus \mathbb{Z}_4$.  I think the first step is to find a basis of $M$. Clearly $(1,0,0)$ and $(0,0,1)$ are in a basis of $M$. I don't know how to find the third element in that basis. Also the basis in $\mathbb{Z} \oplus \mathbb{Z}_4$ are $(1,0)$ and $(0,1)$.
8245
It is suggested as an exercise in Serge Lang's book "Algebra" to show that the commutator subgroup $G^c$ of a group $G$ is a normal subgroup.   I'd like to do that but I am afraid I need help,   I think the first thing I need to figure out is how a general element in the commutator subgroup looks like, so that I can check that the defining condition for normality is satisfied.  That is, supposing for a moment that a general element in $G^c$ is denoted by $g$, I need to show that $aga^{-1} \in G^c,$ for all $a \in G$.  But here I get stuck, first because I am unsure how to write a general element in $G^c$ - a simple product in $G^c$ is of the form $xyx^{-1}y^{-1}aba^{-1}b^{-1}$ where $a,b \in G$. I cannot see a way to simplify this - I am sure there is one, but somehow I am blind today.  The second thing then is, even if one tries out the conjugation of a simple element like $xyx^{-1}y^{-1}$ in $G^c$, again not simplification offers itself easily I think .. what am I missing ?  An alternative would be to find a homomorphism of $G$ whose kernel is precisely $G^c$ - here I tried to think of this as a map $G \times G \to G$ but whatever I cook up is not a homomorphism.  Thanks for your hints !!
41977
Let $G$ be a group of order $56$. (We do NOT assume the Sylow-$7$ subgroup to be normal.) Then either the Sylow-$2$ subgroup is normal or the Sylow-$7$ subgroup is normal. How to prove?  My idea: consider the case $n_2=7,n_7=8$. Then the number of elements in all Sylow $2$ -subgroups and Sylow $7$-subgroups is at most exactly $(7-1)*8+(2-1)*7+1=56$. This cannot give a contradiction. We need to prove that $G$ is simple. Then $|G||(8!/2)$. To prove $G$ simple, we need to prove there does not exist normal subgroups of order $2$ and order $4$. But I do not know how to prove.