Let G be a group. Prove that the function: $G→G, x→x^2$ is a group  homomorphism if and only if G is an abelian group.   Is this the same as if $f:G→G$ is defined by $f(x) = x^2$ ?
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The function $f: G→G$ defined by $f(x) =x^2$ is a homomorphism if and only if $G$ is abelian.  Can anyone give me any tips how to work on this question?
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Let $G$ be a group.  If $G$ is non-trivial, then there exists a non-trivial homomorphism from $\mathbb{Z}\to G$.  How would one prove the above statement?
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Question: Let $G$ be a group. Prove that $Z(G)$ is a subgroup of $G$.   If I want to show that $Z(G)$ is a subgroup of $G$ that means I have to show that it is closed under group operation?  Here is my attempt.   Let $a,b$ be elements in $Z(G)$ and $x$ be an element in $G$. Then $ax=xa$ which is under group multiplication commutative and under inverse $(a^{-1}) x=x(a^{-1})$. And hence $(a^{-1})bx= (a^{-1})xb=x (a^{-1})b$ which is under group operation so $(a^{-1}),b$ are elements in $Z(G)$ thus a subgroup of $G$.  Really appreciate if anyone can help me by directing me if my attempt is not good. Thanks in advance.
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Let $G$ be a group and let $f:G\to G$ be an isomorphism. Let $H = \{ a \in G \; | \; f(a)=a^{-1}\}$ Prove that $H$ is a subgroup of $G$ if and only if $G$ is abelian.  My attempt: I already know that $H$ is a subset of $G$ since $a$ is in $G$. And that $H$ is $a$ group if and only if for every $a$ and $b$ in $H$, $ab^-1$ is in $H$. That is, if $f(a)=a^{-1}$ and $f(b)=b^{-1}$ then $f(ab^-1)=(ab^{-1})^{-1}=(b^{-1})a$.  But I don't know how to derive this nor the converse based on the abelian property of $G$ that for every $a$ and $b$ in $G$, $ab=ba$.  Please help some. Thanks!
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Question:  Suppose that $\alpha: G \to H$ is a surjective homomorphism of groups. Let $U$ be a subgroup of $H$ and prove the following:  If $U$ is normal in $H$, the pre-image of $U$ is normal in $G$.  My answer:  Define $Y = \{g \in G | g^{\alpha} \in U\}$.  $U$ is normal in $H$ so $h^{-1}uh \in U$ for $h \in H, u \in U$.  Now $h = g^\alpha$ and $u = y^\alpha$ for $g \in G, y \in Y$.  So we have $(g^\alpha)^{-1}(y^\alpha)(g^\alpha) \in U$  $\implies (g^{-1}yg)^\alpha \in U$  $\implies g^{-1}yg \in Y$  So $Y$ is normal in $G$.  Is that correct?
82071
If $f:G \rightarrow H$ is group homomorphism, and $H$ is abelian, then $G$ is abelian.  Is that statement correct? Here's my attempt of proof:  Let $a,b \in G$, then: $ab=f^{-1}f(ab)=f^{-1}f(a)f(b)=f^{-1}f(b)f(a)=f^{-1}f(ba)=ba$  Is this correct? I relied on first on $f$ being group homomorphism, then $H$ being abelian.
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Let $f$: $Z_m \rightarrow G$, $f(n)$ = $a^n$. Show that f is a function and an homomorphism.  Doing a review on the whole semester, and a little refresh on this topic would be great.   By definition I know $G$ has the same order of $Z_m$, but not sure how to attack this problem with ease.   I also know that cyclic groups of the same order are isomorphic.   (I cannot use the fact that cyclic groups of the same order are isomorphic, that's why I ask for another approach, because I can't use theorems or lemmas that have not been discussed in class.)
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Let $G$ be a finite group and $f:G\to G$ an isomorphism. If $f$ has no fixed points (i.e., $f(x)=x$ implies $x=e$) and if $f\circ f$ is the identity, then $G$ is abelian. (Hint: Prove that every element in $G$ has the form $x^{-1}\cdot f(x)$.)   With the hint, I can see that for all $t\in G$, $f(t)=t^{-1}$, and since $f$ is an isomorphism, $G$ is abelian. But I can't see how to prove the hint or why it's evident.  Source : Rotman J.J. Introduction to the theory of groups, exercise 1.50
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Let $G$ be a group in which $a^2=e$ for all elements of $a$ of $G$.   Show that $G$ is Abelian.  I need help on this problem.  Appreciated!
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I need to find all group homomorphisms of $S_n\to\mathbb{C}^*$ (for $n\ge2$), and also all group homomorphisms of $A_n\to\mathbb{C}^*$ (for $n\ge2$).  Someone on this site, two years ago, seems to have already asked the same question (). I understand what the questioner says there, but I don't understand it when he says:     Therefore there is exactly one $g:\mathbb{Z}/2\mathbb{Z}\to\mathbb{C}^*$, so all elements of $a\in\mathbb{C}^*$ give us $\operatorname{ord}(a)\mid g[\mathbb{Z}/2\mathbb{Z}]$, so $a$ can only be $-1$ or $1$. Now we know that $f:S_n\to\{-1,1\}$ &hellip;   What is he saying there? And how to go on from there? (I do not understand the answer given either &hellip;)  I hope that someone here can help me further. (And if I'm not allowed to ask this question because is has already been asked two years ago, please tell me what to do instead).  Thank you in advance!
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If $G$ is a group show that if $(a \cdot b)^2 = a^2 \cdot b^2$ then $G$ must be abelian.   $\begin{aligned}(a \cdot b)^2 = a^2 \cdot b^2 &amp; \iff (a\cdot b)\cdot(a \cdot b) = (a \cdot a)\cdot (b \cdot b) \\&amp; \iff a \cdot (b \cdot (a \cdot b)) =a(a \cdot (b\cdot b)) \\&amp; \iff (a^{-1} \cdot a) \cdot (b \cdot (a \cdot b)) =(a^{-1}\cdot a)(a \cdot (b\cdot b))\\&amp; \iff (b \cdot (a \cdot b)) =(a \cdot (b\cdot b)) \\&amp; \iff (b \cdot a) \cdot b =(a \cdot b)\cdot b\\&amp; \iff (b \cdot a) \cdot (b \cdot b^{-1}) =(a \cdot b)\cdot (b \cdot b^{-1}) \\&amp; \iff b \cdot a = a \cdot b\end{aligned}$  Thus $G$ must be abelian. Is this right? Is there less clunky way to write it if it's?
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Let G be a group, where $(ab)^3=(a^3)(b^3)$ and $(ab)^5=(a^5)(b^5)$. Prove that $G$ is abelian group.  Thank you in advance. Any help is appreciated.
103432
Just starting to properly learn the chapter on homomorphisms, and it will be great if I can understand the following question I have:  How could I actually show that when there are two groups $G$ and $H$ (finite), and $|G|$ and $|H|$ are coprime, then then homomorphism $\Phi:G \rightarrow H$ must be trivial.   My idea: Since $Im(\phi)$ and $Ker(\phi)$ are respective subgroups of the groups $G$ and $H$, then by Lagrange's Theorem $|Ker(\phi)|||G|$ and $|Im(\phi)|||H|$. I think if we just show that the |$Im(\phi)$| is $1$, then it implies that the $|Ker(\phi)|$ is equal to the order of $G$. Hence $\phi$ must be trivial. But if this is correct, I cant seem to show $|Im(\phi)|=1$. However, is this the correct way?   Can someone please help me show the homomorphism $\Phi$ is trivial if my idea is completely off?
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Prove or disprove: if $f:G \rightarrow H$ is a surjective group homomorphism, then   1) $f(A)$ normal implies $A$ normal.  2) $S$ is a p-Sylow subgroup of $G$ implies $f(S)$ is a Sylow-p subgroup of $H$.   I believe 1 is false since I believe we need injectivity, but I cannot think of a counterexample...  I believe 2 is true but do not know how to show it...
121422
Let $G_1$ and $G_2$ be groups and $\phi:G_1\to G_2$ be a group homomorphism. Show $\phi(g^{-1})=(\phi(g))^{-1}$ $\forall g_1\in G_1$.  So far I have:  $\phi(e)=e$ by definition, where $e$ is the identity element. Then $\phi(g_1)\phi(g_1^{-1})=\phi(g_1g_1^{-1})=\phi(e)=e$ since $\phi$ is a group homomorphism.  I know I can get $\phi(g^{-1})=(\phi(g))^{-1}$ $\forall g_1\in G_1$ from that, but I don't know HOW.  Any help/hints would be welcomed. ^_^
113212
True or false. (Prove or give a counterexample.) Let $G$ be a group. Then $|g| = |\phi(g)|$, for all homomorphisms $\phi: G \to G$ and all $g \in G$. Solution. False. $\phi: \mathbb Z_{10} \to \mathbb Z_{12}$ defined by $\phi(x)=0$ for all $x \in \mathbb Z_{10}$ is a counterexample. This function is a homomorphism because $\phi(x+y) = 0 = 0+0 = \phi(x) + \phi(y)$ for all $x, y \in \mathbb Z$ (this function is discussed in problem 3 in assignment 7 as the function sending $[1]$ to $[0]$). The order of $g=1$ is infinity. The order of $\phi(1)=0$ is one.  The problem is that he said $G \to G$ is equivalent to $\mathbb Z_{10} \to \mathbb Z_{12}$, which is think is not right. Explain please? EDIT: Please look at this test and give me your honest opinion,
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Let $f:G\to K$ be a surjective group homomorphism and $H$ be a subgroup of $G$ satisfying $N_G(H)=H$. Then $N_K(f(H))=f(H)$. I'm trying to prove or disprove it. perhaps, this proposition is true. But, how can we prove it? Edit: I thought it is true because WIKI said it is true without proof