prove that if $k|m$, then $\mathbb{Z}_m$ has a subgroup of order $k$.  im not sure where to start with this.  any help is appreciated.  thanks!
124218
Prove that if $k\mid m$, then $Z_m$ has a subgroup of order $k.$  Ok, so this doesn't look like too hard of a problem. So do I just show that it is closed under multiplication and inverse? I just don't know how I would do that with a problem like this. Can anyone help me get started please?
87134
How calculate all subgroups of $(Z_{12}, +)$?  I know that the order of subgroups divide the order of the group, but there is such a smart way to calculate the subgroups of order 6?
7587
Let $H$ and $K$ be subgroups of a group $G$, and let $HK=\{hk: h \in H, k \in K\}$, $KH=\{kh: k \in K, h \in H\}$. How can we prove that $HK$ is a subgroup iff $HK=KH$?
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.
32431
Let $G$ be a group and $H$ be a normal subgroup of index $p$ ( a prime ) ; suppose $K$ is a subgroup of $G$ not contained in $H$ , then is it true that $G=HK$ ? ( I know that the fact is true if $p=2$ , but I don't know for the general case . Please help . Thanks in advance )
841
I don't know about the Sylow Theorems.  But I have been wondering about a proof of the fact that a group or order $pq$ where $p$ and $q$ are distinct primes must be cyclic. I can't quite work out the details, but here is the general idea. I would like help with filling in details. I assume that it is already known that $G$ has subgroup(s) of order $p$ and subgroup(s) of order $q$.  If $G$ is a group of order $pq$ ($p\neq q$), then I know that $G$ has a subgroup $H$ of order $p$ and a subgroup $K$ of order $q$. Then $H\simeq \mathbb{Z}_p$ and $K\simeq \mathbb{Z}_q$.   But then $H\oplus K \simeq \mathbb{Z}_{pq}$, so I would think that $H\oplus K \simeq G$. I guess one could do an internal direct product instead of an external direct product, but I don't know that $H$ and $K$ are normal subgroups.  I am asking for help completing this argument.  Edit: I see from the comments below that I might need to assume that the smaller prime does not divide the larger prime minus $1$. Or maybe it is enough to assume that the primes are greater than or equal to $3$ (Still distinct).
34966
I calculated the subgroups of $\mathbb Z / 2 \mathbb Z \times \mathbb Z / 2 \mathbb Z$ as $\{ (0, 0 ) \}, \{ (0, 0 ), ( 1, 0 ) \}, \{ (0, 0 ), (0, 1) \}, \{ (0, 0 ), (1, 1) \}$ and $ \mathbb Z / 2 \mathbb Z \times \mathbb Z / 2 \mathbb Z$ and itself. I  a bit and found couple of answers which did not include $\{ (0, 0), (1, 1)\}$. Isn't $\{ (0,0), (1,1)\}$ subgroup of  $\mathbb Z / 2 \mathbb Z \times \mathbb Z / 2 \mathbb Z$ ?
109849
Show that cyclic group of finite order $n$ is isomorphic to $(\mathbb Z_n,+)$ Actually, the answer to this question is available in MSE but I am not getting it,and one answer is not in the English language.Please kindly give a proof in easy language. Thank you.\ Here is a solution I don't understand:
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.
18013
I am participating in an Introductory course in groups and I have the following question:     Let $G$ be a finite group of order $8$. Prove that $G$ has a subgroup of order $4$ and a subgroup of order $2$.   i know how to prove that $G$ has a subgroup of order $2$ but i can't show the existence of a subgroup of order 4. I am only a begginer in groups so please take this into account.
43020
I am wondering how to prove the formula $|HK| = \dfrac{|H| \; |K|}{|H \cap K|}$ where $H$ and $K$ are finite groups. Let $|H|$ denote the order of $H$ and so on.  I have seen proofs from these websites:  and .  But I do not really understand the idea of the proof. Can someone explain the proof to me, please? Thanks so much.
29223
I'm a bit lost on this problem:      Show that a group of order $p^2$ is isomorphic to $\mathbb{Z}_{p^2}$ or $\mathbb{Z}_{p}$ $\times$ $\mathbb{Z}_{p}$.    The only thing I can think of that may relate to this problem is Lagrange's Theorem, where the order of a subgroup divides the order of a group. But I'm not sure if I can even use that for this problem... Any help would be tremendously helpful. Thank you!
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!
76898
Let $m$ and $n$ be two nonnegative integers. Assume that there is a group isomorphism $\mathbb{Z}^m \cong \mathbb{Z}^n$. Prove that $m = n$.   I tried using a contrapositive, ($m \neq n$ implies $\mathbb{Z}^m \ncong \mathbb{Z}^n$), and I think the problem is that there won't be a homomorphism, but did not get anywhere. Is there a better approach to this problem?
39404
I dont understand the following very simple statement:  If $\tau \in S_n$ has order $m$, then $\sigma \tau\sigma^{-1}$ has also order $m$.  The proof is:  Suppose $\tau$ has order $m$.   $(\sigma \tau \sigma^{-1})^{m} = \sigma \tau^{m}\sigma^{-1}= (1)$  Then (using  exercise 17) they say that the order of $\sigma\tau\sigma^{-1}$ cannot be less than $n$ (remember we are in $S_n$), since $(\sigma\tau\sigma^{-1})^k=1 $ implies $\sigma\tau^k\sigma^{-1} =(1)$, and then $\tau^k=\sigma^{-1} \sigma = (1)$  I don't understand the last part.... Who can help me?
22464
I would be very thankful if someone could give me a hint with proving this. It's a very common exercise in abstract algebra textbooks.     If $G$ is a group with a subgroup $H$ of finite index $n$, then $G$ has a normal subgroup $K$ contained in $H$ whose index in $G$ is finite and divides $n!$.   I found the proof on  at some point (although the proof appears to be no longer there), but I got lost in one of the details.
27813
I've been reading some stuff about algebra in my free time, and I think I understand most of the stuff but I'm having trouble with the exercises. Specifically, the following:     Prove that a nonempty subset $H$ of a group $G$ is a subgroup if for   all $x, y \in H$, the element $xy^{-1}$ is also in H.   Proving that the identity is in $H$ is easy: just take $x=y$, so $x x^{-1} = 1 \in H$. However, I'm having trouble showing that multiplication is closed and that each element in $H$ has an inverse. Can anyone give some hints?