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May
8
answered Suppose $G = G_1 * G_2$. let $c \in G$ and let $A = cG_1c^{-1}$. Show that $A\cap G_2 = \{1\}$
May
7
comment Abelianised dihedral group isomorphism
Suppose that $n =2k$. You have $x^2 = 1$ and $x^{2k}=1$. But $x^{2k}=1$ follows from $x^2=1$ because $x^{2k} = (x^2)^k$.
May
7
answered Abelianised dihedral group isomorphism
May
1
comment Implications of the order of subgroups, given the order of the parent is a composite number
Yes, if the $x_i$ are all primes; this is essentially Cauchy's theorem.
Apr
30
comment “Quotient” as a verb
People are just verbing the word "quotient".
Apr
29
answered Finitely presentated subgroups of a group are normal?
Apr
29
comment Distinct four elements in G
It's not true. The cyclic group of order $5$ has exactly four (non-trivial) elements whose fifth power is the identity, and they are certainly distinct.
Apr
29
comment What is the easiest way to generate $\mathrm{GL}(n,\mathbb Z)$?
@Leon. Diagonal matrices are elementary matrices.
Apr
23
comment Generate specific reduced words that “violate freeness”
I'm not sure I understand the problem. What hypothesis do you think you need about the order of $g_1$? If everything collapses, I think you're going to contradict your assumption that $g_1$ and $g_2$ do not commute.
Apr
23
comment Generate specific reduced words that “violate freeness”
Yes, for (1), you can conjugate the word by $g_{i(1)}^{m(1)}$ to get the first and last syllables to be distinct. I don't yet understand the second question, though.
Apr
23
comment Trying to show $|ab|$ divides lcm$(|a|,|b|)$
Try looking at the set of distinct (cyclic) subgroups of order $p$.
Apr
23
comment Paradoxical Decomposition
@AllAboutGroups We can write $F_2 = B\mathscr{F}(B^{-1})\cup\mathscr{F}(B)$, and $F_2 = A\mathscr{F}(A^{-1})\cup\mathscr{F}(A)$ where, for $X\in\{A,B,A^{-1},B^{-1}\}$, we define $\mathscr{F}(X)$ to be the set of reduced words in $F_2$ beginning with $X$.
Apr
20
answered Paradoxical Decomposition
Apr
20
comment Is $A_4$ isomorphic to $D_3\times \mathbb Z_2$?
Yes, $A_4$ has no element of order $6$, while $D_3\times Z_2$ does. You can use the same idea with the second question: what are the orders of the elements of $H$ and of $Z_4$?
Apr
16
comment Semidirect product.
Since $z$ has order $2$, you'll need the relation $z^2 = e$. But you also need relations describing the action of $z$ on the direct product of the dihedral groups: $z^{-1}x_1z = x_2, z^{-1}y_1z = y_2$.
Apr
13
comment Internal Direct Product
Well, you're almost there. You've shown that your subgroups $H$ and $K$ must be abelian, so what can you say about their direct product?
Apr
10
answered Are free products of finite cyclic groups perfect?
Apr
10
comment Are free products of finite cyclic groups perfect?
It's not perfect. There is an epimorphism onto the cyclic group of order $6$, which is $\mathbb{Z}_2\oplus\mathbb{Z}_3$.
Apr
10
comment Find all sylow subgroups of$ D_{2p^k}$, where p is prime
If $p=2$, then $D_{2p^k}$ is a $2$-group, meaning that it has only a Sylow $2$-subgroup equal to itself.
Apr
10
answered If $f(g) = g^k$ is a homomorphism on a finite group $G$ and $k < |G|$ does not divide $|G|$, must $G$ be abelian?