By Gauss' Lemma on a property of primes, if $p\mid ab$, then either $p\mid a$ or $p\mid b$. (Why?)
Hint:
Now the contrapositive states that if both $p\nmid a$ and $p\nmid b$, then $p\nmid ab$.
Since $\gcd(3,4)=1$, we have $\Bbb Z_3\times \Bbb Z_4\cong \Bbb Z_{12}$ (by the Chinese Remainder Theorem), so the group is cyclic and so, hence, all its subgroups are cyclic; it is then a matter of finding with elements of $\Bbb Z_{12}$ generate a proper subgroup then seeing which of these is di...
Working on the hint by @HenningMakholm . . .
Each element of $\Bbb Z_2\times\Bbb Z_{36}\times\Bbb Z_{10}$ is a triplet $([a]_2, [b]_{36}, [c]_{10})$ and its order is $\operatorname{lcm}(A, B, C)$, where $A$ is the order of $[a]_2$ in $\Bbb Z_2$, $B$ is the order of $[b]_{36}$ in $\Bbb Z_{36}$, a...
Hint: $$G/Z(G)\cong\operatorname{Inn}(G),$$
where $\operatorname{Inn}(G)$ is the group of inner automorphism of $G$ under composition of functions.
Reference: Theorem 9.4 of Gallian's "Contemporary Abstract Algebra (Eight Edition)"
This holds because a flip followed by a rotation (or vice versa) is, again, a flip, so has order two. Don't be afraid to use your geometric intuition here!
And what about $(FR)^{-1}=R^{-1}F^{-1}$? It just means $R^{-1}F^{-1}=FR$ in this context.
The infinite cyclic group is $\Bbb Z$. It has $$\langle c\mid\rangle$$ as a presentation, since it is free.
If we introduce the relator $c^n$, then we get the presentation $$\langle c\mid c^n\rangle,\tag{$\mathcal{P}$}$$ which defines $\Bbb Z_n$.
But we get from the presentation of $\Bbb Z$ abo...
1) No. One needs to show that the product (so, here, addition) of any two elements of $K$ is in $K$.
2) No. One usually says that associativity in $K$ is inherited from $\Bbb Z$, since, indeed, all elements of $K$ are integers and the set of integers under addition is associative.
3) Correct!
...
Since $$M_2^n=
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & n \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}$$
by induction on $n$ for $n\in\Bbb Z$, we have that the subgroup $\langle M_2\rangle $ generated by $M_2$ is isomorphic to the free group $\Bbb Z$.
A word on notation: $6\Bbb Z=\{6z\mid z\in \Bbb Z\}$ and is usually taken to be the additive group $(6\Bbb Z, +)$ under addition.
To show the two groups are isomorphic, consider $$
\begin{align}
\varphi: (\Bbb Z,+)&\to(6\Bbb Z, +) \\
z&\mapsto 6z.
\end{align}
$$
Then, for any $x, y\in\Bbb Z$, w...
Yes, your proof is correct.
You started, essentially, with $a, b\in H$, then showed that $[a, b]\in H$. This works because it shows that all the generators of $G'$ are in $H$ and $H\lhd G$.
You're correct.
You are given that the group is abelian. Multiplicatively, the group is given by the presentation
$$\langle a,b,c\mid a^6b^9c^6, a^8b^{12}c^4\rangle^{\operatorname{ab}}. \tag{$P$}$$
What you have done is introduce $t=abc, u=b, v=c$ to $(P)$ via Tietze transformations to get th...
Since the group isomorphism problem is undecidable (which is to say that there exists no Turing machine that will halt in finite time, given two groups in a suitable code, whether the two groups are isomorphic), there is no algorithm for solving this problem: Take two groups, give them each a def...
