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Why is $\Bbb Z/4\Bbb Z\times \Bbb Z/12\Bbb Z\times \Bbb Z/40\Bbb Z$ not isomorphic to $\Bbb Z/8\Bbb Z\times \Bbb Z/10\Bbb Z\times \Bbb Z/24\Bbb Z$?

We have that the first group is isomorphic to $$\Bbb Z_4\times(\Bbb Z_3\times \Bbb Z_4)\times (\Bbb Z_5\times \Bbb Z_8),$$ while the second group is isomorphic to $$\Bbb Z_8\times(\Bbb Z_2\times \Bbb ...
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5 votes

Why is $\Bbb Z/4\Bbb Z\times \Bbb Z/12\Bbb Z\times \Bbb Z/40\Bbb Z$ not isomorphic to $\Bbb Z/8\Bbb Z\times \Bbb Z/10\Bbb Z\times \Bbb Z/24\Bbb Z$?

One simple way to see this is that the second group has an element $(0,5,0)$ of order $2$ and no element $g$ such that $g+g=(0,5,0).$ There is no element like this in the first group. If $(x,y,z)\in\...
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3 votes
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How to determine the invariant factors of an abelian group of order $2^6 \cdot 3^5 \cdot 5^7$

There is a fact that serves to find them algorithmically. Fact. Let $G$ be a finitely generated group, then there exists $m\geq 0$, $p_1,\ldots,p_n\in \mathbb{Z}_{+}$, distinct primes and naturals \...
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4 votes
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If both finite groups $G_1$ and $G_2$ (of equal order) possess equal number of elements of highest possible order, will $G_1\simeq G_2$ hold?

No, e.g. $\Bbb Z_8\oplus\Bbb Z_4$ and $\Bbb Z_8\oplus\Bbb Z_2\oplus\Bbb Z_2$ have an equal number of elements of order $8$.
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On a proof that the quaternion $8$-group $Q$ mod its center is isomorphic to the Klein four group

By Lagrange's Theorem, $$\begin{align} |Q/Z(Q)|&=|Q|/|Z(Q)|\\ &=8/2\\ &=4. \end{align}$$ As noted in the comments, there is only two groups of order four up to isomorphism: the Klein four ...
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5 votes
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If two finite Abelian groups posses equal number of elements of some particular order, then justify they are isomorphic.

Trivial case: In fact any group has a unique element of order $1$, i.e., $\Bbb{Z}_4$ and $K_4$ has same number of element of order $1$. Consider $\Bbb{Z}_2\times\Bbb{Z}_6$ and $\Bbb{Z}_{12}$. The ...
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Let $\phi:\Bbb{Z}_{20} \to \Bbb{Z}_{20}$ be an automorphism and $\phi(5)=5$. What are the possibilities of $\phi(x)$?

We know, an isomorphism maps generators to generators. Gen Z_20 = {1,3,7,9,11,13,17,19}. These are the possible competitors for automorphisms. The other condition according to question is Φ(5)=5, say ...
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2 votes
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Finding order of elements order-wise in product of groups.

The direct product of two abelian groups is abelian. Thus $$G= C_{8}\times C_2\times C_1\times C_1$$ is abelian. However, since $Q_8$ is nonabelian, $Q_8\times A$ is nonabelian for any group $A$. Thus ...
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What is the role of $\gcd(n,|G|) = 1$ in the following question?

Addressing injectivity. I am clueless about the logarithm business. Verify that $a\mapsto na$ is a morphism. Suppose $|G| =m$, $\mathrm{gcd}(m,n)=1$ and $na=0$ for some $a\in G$. Pick $r$ such that $...
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What is the role of $\gcd(n,|G|) = 1$ in the following question?

The logarithm is a function from real numbers to real numbers. You can't just feed it elements of some arbitrary group without doing some work to try to define what it means first.
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2 votes

Find isomorphisms between two small finite groups $G, H$ of equal cardinality.

Since there are six elements in each underlying set and $6=2\times 3$, we can use the following theorem: Theorem: Let $p$ be an odd prime. Then each group of order $2p$ is isomorphic to either $\Bbb ...
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Show that $4ℤ/12ℤ ≃ ℤ_3$

One possibility to solving your problem would be by constructing an isomorphism between the two groups. For example consider: $$ \varphi: 4\mathbb{Z} /12\mathbb{Z} \to \mathbb{Z} / 3 \mathbb{Z} $$ $$ ...
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4 votes
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Find pair of product of four groups that has the same order, but not isomorphic.

So,$ G_1=C_{16}×C_1×C_1×C_1$ and $H=C_8×C_8×C_1×C_1 $will not help; as then will have equal orders as well as Isomorphism too. Both groups neither have the equal order nor they are isomorphic. $G_1$ ...
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2 votes

Find pair of product of four groups that has the same order, but not isomorphic.

How about $C_2 \times C_4 \times C_8 \times C_{16}$ and $C_2 \times C_4 \times Q_8 \times C_{16}$? The first group is abelian, the second is not.
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Prove a particular subset of a group is closed under inverse

You can essentially do it the second way, withan adjustment: you have that $g\in B_a\implies ga=ag\implies a^{-1}g=ga^{-1}$. That's not too hard to get. Now, we need to prove $g^{-1}a\in A$. So ...
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1 vote
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Find isomorphisms between two small finite groups $G, H$ of equal cardinality.

Note $G$ and $H$ are cyclic groups with order $6$ For $G$, $e=10$, $G=\langle 2\rangle$, $~~~2^1=2,~2^2=4,~2^3=8,~2^4=16,~2^5=14,~2^6=10=e$ For $H$, $e=15$, $H=\langle 3\rangle$, $~~~3^1=3,~3^2=9,~3^3=...
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1 vote

A "different" third isomorphism theorem

The best thing you can say is that there is a surjective map $G/K\to G/H$ whose kernel is $H/K$. So instead of cokernels you take kernels.
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Homomorphism between two groups

Firstly, as was mentioned in the comments, the statement "$G/\text{ker} \phi \cong H$" has no meaning without first defining $\phi$. The proper question to ask would be if $G/N\cong H$ for ...
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7 votes
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Morphism of free groups that induces isomorphism on abelianizations

The alternating group $A_5$ is generated by $a = (12345)$ and $b = (12)(34)$. These give a transitive action of the free group $F_2$ on $ \{ 1, 2, 3, 4, 5 \}$. The stabilizer of $5$ is a subgroup $H$ ...
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2 votes
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Left action and group automorphism

To esentialy sum up and promote the comment stream to the question itself to an answer: Let the group under discussion be denoted by $G$. The left (or right) action by $a \in G$ given by left (right) ...
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2 votes
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For which groups can an isomorphism between subgroups always be lifted to an automorphism?

For groups of odd order with your property for groups of order $p$ only, so a larger class of groups, this paper from 1973 gives a structural result. They are pretty close to direct products of $p$-...
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1 vote

Show $S_7$ is isomorphic to the subgroup of all those elements of $S_8$ which leave the number $5$ fixed

For every $a\in A:=\{1,\dots,n\}$, set $B:=A\setminus\{a\}$. Then, $\operatorname{Stab}(a)\stackrel{\varphi}{\cong} S_B$ via $\sigma\mapsto\varphi(\sigma):=\sigma_{|B}$. In fact, firstly, $\sigma_{|B}\...
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