# Tag Info

## Hot answers tagged group-theory

4

A really simple example is inside the group $\mathbb Z/5\mathbb Z$. $2$ has order $5$ since it generates: $2, 4, 1, 3, 0$. but $2^2 = 4$ and it also has order $5$ since it generates: $4, 3, 2, 1, 0$.

4

Its not true. The existence of a homomorphism shows nothing, since there is always the trivial homomorphism. For instance there is bijection $\mathbb Z \to \mathbb Q$, and also a homomorphism $\mathbb Z \to \mathbb Q$ (one for any rational number...), but they are not isomorphic.

3

It's not clear by your armgument, why there must exists a $\sigma$ for each $m$ such that $\sigma(m)=k$. On the other hand, there's no need for a counting argument. Given, $1\le m<k\le n$, let $\sigma$ be the transposition $(mk)$. That is, the bijection which switches $m$ and $k$ and leaves every other number fixed.

3

Try the following commands: G:=SmallGroup(m,n); H:=Image(IsomorphismFpGroup(G)); RelatorsOfFpGroup(H);

2

No, this will not work. As a counterexample, consider $A,B$ two different finite groups with the same number of elements (e.g. $A=\mathbb{Z}_2\times\mathbb{Z}_2$ and $B=\mathbb{Z}_4$), choose any bijection $\phi_1$ and take the trivial homomorphism $\phi_2(a) = 1_B$.

2

Hint : how many transpositions have you in $S_5$? how many double transpositions have you in $S_5$?

2

For the first part, consider the additive group $\mathbb{Z}/5\mathbb{Z}$ and pick appropriate elements (try 1 and 2). For the second, I'm not sure there's an elementary way to do it other than manually counting, but for a prime $p$, I believe the number of primitive roots (elements that generate the cyclic group) is $\phi(\phi(p))$, where $\phi$ denotes ...

2

The key result for both questions is If $ord(x)=n$, then $ord(x^k) = \dfrac{n}{(n,k)}$ $ord(x^2)=5$ implies $n=5(n,2)$ and so $n=5$ or $n=10$. Both cases are possible. If you know that $\mathbb Z_{17}^{\times}$ is cyclic of order $16$, then there are as many generators as $k$ such that $0\le k \le 15$, $(16,k)=1$ and so there are $\phi(16)=8$ ...

2

Hint: Suppose we are given an integer $x$, and we wish to determine whether $\overline{x} \in \mathbb{Z}/n\mathbb{Z}^\times$ for some $n \in \mathbb{Z}$. If it were, then we could find an integer $y$ such that $xy \equiv 1 \pmod{n}$. In other words, $n$ divides $xy - 1 \implies nm = xy - 1$ for some $m \in \mathbb{Z}$. Rearranging the above yields $xy - ... 1$o(a)$is equal to$o(\langle a\rangle)=o(H).$But,$o(ab)$is not necessarily the order of$\langle a\rangle.\langle b\rangle=HK$. You may track which argument is not valid in your proof. The statement in the title is wrong. It should be If$o(a)=m$,$o(b)=n$and if$ab=ba$then$o(ab)$divides$lcm(o(a),o(b))$. 1 Your argument is correct. A simpler and better way is as follows: $$N_i/N_{i-1} \leq Z(G/N_{i-1}) \Longleftrightarrow N_i/N_{i-1} \mbox{ commutes with } G/N_{i-1} \Longleftrightarrow [N_i/N_{i-1}, G/N_{i-1}]=1 \Longleftrightarrow [N_i,G]\leq N_{i-1}$$ By homomorphism$\phi$, we have$\phi[x,y]=[\phi(x),\phi(y)]$. Hence,$$[M_i, H]=[\phi(N_i), ... 1 Indeed, as per the comment above, an automorphism preserves the order of elements, so all you can say is that a transposition must be sent to an element of order two, so namely either another transposition or an element of$S_5$that is written as the product of two disjoint transpositions (why are these the only elements of order 2 in$S_5$?). But here's ... 1 Hint: Show that every homomorphism$g:\mathbb{Z}_p \to \mathbb{Z}_{p^2}$factors through the homomorphism$k:\mathbb{Z}_p \to \mathbb{Z}_{p^2}$defined by$k([x])=[-px]$(i.e there exists$\bar g: \mathbb{Z}_p \to \mathbb{Z}_p$such that$k\bar g=g$). To do so note that$k$is an isomorphism onto the kernel of$q:\mathbb{Z}_{p^2}\to\mathbb{Z}_{p^2}\$ defined ...

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