The question is to work out the number of semi-direct products for from Q to H, where   $$ H = C_{42} , Q = C_{3} $$  I did:  $Aut(C_{42}) = C_2 \times C_6 = C_{12}$  3 divides 12 telling us that they're are some semidirect products.   If we let Q = (1, 2, 3). We can see that they're are 3 elements in Q (1, 2 and 3) that divide 12, thus there are 3 semi direct products.  Is this correct reasoning?
123632
There is a semi direct product if you go $\theta: Q \rightarrow Aut(H)$.  $H = C_{17}, Q = C_2$.  $Aut(H)\cong C_{16}$.  From here, how do I construct the semi direct products?   I said, in $C_{16}$, there are two elements of order that divide $2$, which are $1$ and $2$. Therefore there are two semi direct products. Also $17 \times 2 = 34$ tells me that in both semi direct products there will be $34$ elements.  How do I construct the semi direct products from here though?
68048
We have to prove that  "the product of two numbers ending in 6 also ends with 6" mathematically.  I have no clue how to start. I don't want you to prove it for me! but some hints would be very helpful as I'm totally stuck on how to start this prove.
68220
I would first like to make sure of the following concepts, note that all groups are abelian. Let $A,B$ be subgroups of $G$.  (1) Cartesian product (a.k.a. external product): The Cartesian product of subgroups $A$ and $B$ is $A \times B = \left\{ (a,b) \:{:}\: a \in A, b \in B \right\}$ where the operation by components is the group operation.  (2) Internal product: The internal product of subgroups $A$ and $B$ is $AB = \left\{ ab \:{:}\: a \in A, b \in B \right\}$.  (3) Sum: The sum of subgroups A and B is $A + B = \left\{ a + b \:{:}\: a \in A, b \in B \right\}$.  So (2) and (3) are basically the same objects because (2) is used under the multiplicative notation and (3) is used under the additive notation.  (4) Direct product: The internal product $AB$ is called a direct product and denoted $A \times B$ (confusingly enough) if $A \cap B = \left\{ 1 \right\}$.  (5) Direct sum: The sum $A + B$ is called a direct sum and denoted $A \oplus B$ if $A \cap B = \left\{ 0 \right\}$.  So (4) and (5) are basically the same objects because (4) is used under the multiplicative notation and (5) is used under the additive notation.  I am aware that (1) and (4) (or (5)) are isomorphic and often used interchangeably. When the direct sum is used in a correct way, it causes no confusion at all such as $\mathbb{Z}_{15} = \left\{ 0,5,10 \right\} \oplus \left\{ 0,3,6,9,12 \right\}$.  But when it is used in a rather ambiguous way such as $\mathbb{Z}_2 \oplus \mathbb{Z}_2$, which should be $\left\{ 0+0,0+1,1+0,1+1 \right\} = \left\{ 0,1 \right\}$, which is of order $2$, but I have seen my lecturer saying it is of order $4$, so he must have interpreted it as the Cartesian product owing to the isomorphic nature.  I am really not happy with this as the definition of sum and direct sum clearly states otherwise. Could anyone shed some enlightening lights on this issue?
87460
I want to show rigorously that this is 2. I'm sure there's a faster way than by trying to see if $\sqrt{5} = a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6}$ (linear combination of the basis elements for $\mathbb{Q}(\sqrt{3}, \sqrt{2})$). I'm just not sure what.
69810
I have solved the simpler case of this question (with just $a$ and $b$ and a $b$ length block), but no answer on this site has ever properly answered the generalisation:  Let $a,b,c \in \Bbb{N}.\space $Suppose $a&lt;b&lt;c$. Does every block of $c$ consecutive naturals contain three distinct naturals whose product is divisible by $abc$?  I would be very grateful for any help.
89718
Show $\operatorname{Aut}(C_2 \times C_2)$ is isomorphic to $D_6$ (the group with $x^3=1$, $y^2=1$ and $xy=yx^2$).  I'm not really sure how to express the elements of $\operatorname{Aut}(C_2 \times C_2)$. Would it be sufficient to show the elements of $\operatorname{Aut}(C_2 \times C_2)$, find their order and show they bijectively map to every element of $D_6$ and satisfy $xy=yx^2$?
70171
So I have these parts to this problem:  If $n \in \mathbb N$, then let $\tau(n)$ be the number of positive divisors of $n$. For example, $\tau(6) = 4$.  (a) If $n= p_1^{a_1}p_2^{a_2}\ldots p_k^{a_k}$ where $p_i$ are distinct primes and $a_i\in \mathbb N$, then prove that the number of positive divisors of $\tau(n)=(a_1+1)(a_2+1)\ldots(a_k+1)$.  (b) Find the number of positive divisors of $1,\!633,\!500$.  (c) Prove or disprove that $\tau(nm) = \tau(n)\tau(m)$ for all $m$, $n\in \mathbb N$.  I'm not sure how to prove part a. For part b, I know I can just do a series of divisions to find that answer, so I got $2^2\times 5^3\times 3^3\times11^2$ for the divisors, so I think $\tau(1,\!633,\!500)$ would be 4 in this case. For part c, I think I can do a counterexample of 4 and 6, where $\tau(4)=3$, and $\tau(6)=4$, but $\tau(24)$ is not equal to 12, which is $\tau(4)\tau(6)$, if I did my math right.  So, can anyone help me on part a and tell me if I am doing b and c wrong?
61678
Here are my thoughts so far:    $C_p = \langle x \mid x^p =e\rangle$ $C_q = \langle y \mid y^q =e\rangle$ $C_p \times C_q$ has elements of the form $(x^a,y^b)$ There are $p$ possible values for $x^a$ and $q$ possible values for $y^b$. So there are $pq$ possible elements in $C_p \times C_q$. $C_{pq} = \langle z\mid z^{pq} = e\rangle$ and there are $pq$ elements in this group. For an isomorphism from $C_p \times C_q$ to $C_{pq}$ we send $(e,e)$ to $e$ to ensure that the identities are mapped to each other. However, I'm not sure how to define an isomorphism for the other elements.
97344
Prove that $S_3\times \mathbb{Z}_2$ is isomorphic to $D_6$.  My attempt:  I tried to decompose $D_6$ as the internal direct product of subgroups $H$ and $K$ isomorphic to $S_3$ and $\mathbb{Z}_2$ respectively:  Let $H=\{1, r^2, r^4, s, r^2s, r^4s\}$  Let $K=\{1, r^3s\}$  $HK=D_6$  $H\cap K=\{1\}$  However, taking $h=r^2$ and $k=r^3s$, \begin{equation*} \begin{split} hk&amp;=r^2\cdot r^3s \\ &amp;=r^5s \end{split} \end{equation*} \begin{equation*} \begin{split} kh&amp;=r^3s\cdot r^2 \\ &amp;=r^3sr^2ss \\ &amp;=r^3r^{-2}s \\ &amp;= rs \end{split} \end{equation*} $hk\neq kh$  So, this doesn't work. How do I solve this problem?
38542
The function $d(n)$ gives the number of positive divisors of $n$, including n itself. So for example, $d(25) = 3$, because $25$ has three divisors: $1$, $5$, and $25$.  So how do I prove that the product of all of the positive divisors of $n$ (including $n$ itself) is $n^{\frac{d(n)}{2}}$.  For example, the divisors of $12$ are $1$, $2$, $3$, $4$, $6$, and $12$. $d(12)$ is $6$, and $1 · 2 · 3 · 4 · 6 · 12 = 1728 = 12^3 = 12^{\frac{6}{2}} = 12^{\frac{d(n)}{2}}$
9060
I'm having some difficulties continuing this problem. I get that $b^2\mid a$ and $c\mid a$ but I am not sure where to go from there.
44730
I am trying to learn how to find the product of non-disjoint cycles, as you may have guessed from the title. I have the basic idea, but I do not understand it entirely.  I am trying to find $(1352)(256)$ and $(1634)(1352)$, multiplying from right to left.  Please give me some hints/suggestions.  Thank you.
58186
Help me understand problems of this type a bit more intuitively.   The solution $C(3+11−1,11)$ seems simple enough, but I got stuck thinking about how many integers you are choosing from within $x_1$, $x_2$, and $x_3$ and I was trying to get an $n$ - as in $C(n+r-1,r)$ - this way. Also, I though the $r$ was 3 since you have $x_1$, $x_2$, $x_3$ choices. Basically, I went wrong at every step. Help!
106398
I am not sure I understand this I just to an explanation to this question:  If $a, b, c \in \Bbb{Z}^+$ and $a\mid bc$, does it follow that $a\mid b$ or $a\mid c$?
119233
I have already tried with $S_3$, and indeed, the product is $(13)$, and $(13)^2=e$    But what about this: I define + in this way:$45=2$,$26=3$, $1$ is the identity. therefore, $123456=2326=233=3$, and as you can see from the picture, the inverse of $3$ is not itself, thus contradicts the statement.  My question: What is wrong with my argument?  And How do you prove the statement? it seems that my set of numbers perfectly forms a group under the operation I defined via the table above. first it's closed, second there is an identity, third every elements have an inverse. By the way it's commutitive. I tried (45)2=22=6=42=4(52), (45)6=26=3=43=4(56), so associativity seems to be satisfied.
61280
If you toss a coin $n$ times, there are $2^n$ possible sequences of heads and tails. Let $E_n$ be the set of sequences which do NOT contain two consecutive heads and $e_n$, the number of sequences in $E_n$.  Thus $E_3$ $=$ $\{$ $TTT$, $TTH$, $THT$, $HTT$, $HTH$ $\}$ and $e_3$ $=$ $3$.  How many elements of $E_n$ have $T$ as the first toss?  Please can anyone help me out here?
82770
Let $\times$ = cross product and $a$ and $b$ have vector signs on them.  $a = [a_1, a_2, a_3]$  $b = [b_1, b_2, b_3]$  How would I prove this? Do I expand out cross product?  Any help is appreciated!